What is relationship between scalar and vector product?












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I am intrested in relationship between scalar and vector product in $mathbb{R}^3$; I am going to give definitions which I will use in my question.



Scalar product - function $cdot:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})cdot vec{w} = alpha(vec{u}cdot vec{w})+beta(vec{v}cdot vec{w})$



b) $vec{u}cdotvec{v}=vec{v}cdotvec{u}$



c) $vec{u}cdot vec{u} geq 0$ and $vec{u}cdotvec{u}=0 iff vec{u}=vec{0}$



Vector product - function $times:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}^3$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})times vec{w} = alpha(vec{u}timesvec{w})+beta(vec{v}times vec{w})$



b) $vec{u} times vec{v} = -vec{v} times vec{u}$



c) $ (vec{u} times vec{v}) times vec{w} +(vec{v} times vec{w}) times vec{u}+(vec{w} times vec{u}) times vec{v}=vec{0}$



Question: We have endowed $mathbb{R}^3$ with these two structures (structure of scalar product and structure of vector product). I know that space endowed with dot product is called inner space, is there a name for a vector space endowed with vector product? Are vector product and scalar product (as I have defined them) connected structures? Can we somehow exspress vecotor prodcut via scalar product or slacar product via vector prodcut? Is there some formula which can reduce on product to another? What preciasly is connection between these two structures, and can that connection be generalized to $mathbb{R}^n$ (I do not know whether this question makes sense)?



Edit: I am hoping that if there is some connection between these two strcutures that answers not just say it, but rather explain why does this connection hold.



Thank you for any help.










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$endgroup$








  • 1




    $begingroup$
    Maybe the concept of exterior algebra is the genralization you are looking for.
    $endgroup$
    – G.F
    5 hours ago










  • $begingroup$
    The properties listed for $times$ do hold, but they don't make for a satisfactory definition. After all, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ satisfies all three.
    $endgroup$
    – Travis
    5 hours ago










  • $begingroup$
    I think it's rather the Lie algebras.
    $endgroup$
    – Berci
    5 hours ago










  • $begingroup$
    Why has nobody mentioned the geometric product? That combines the dot and cross products, and it generalizes to any number of dimensions. The dot product is the symmetric part, and the cross product is (the dual of) the antisymmetric part, of the geometric product. $$acdot b=frac{ab+ba}{2}$$ $$atimes b=Ifrac{ab-ba}{2}=I(awedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra
    $endgroup$
    – mr_e_man
    1 hour ago












  • $begingroup$
    @mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it?
    $endgroup$
    – Thom
    1 hour ago
















3












$begingroup$


I am intrested in relationship between scalar and vector product in $mathbb{R}^3$; I am going to give definitions which I will use in my question.



Scalar product - function $cdot:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})cdot vec{w} = alpha(vec{u}cdot vec{w})+beta(vec{v}cdot vec{w})$



b) $vec{u}cdotvec{v}=vec{v}cdotvec{u}$



c) $vec{u}cdot vec{u} geq 0$ and $vec{u}cdotvec{u}=0 iff vec{u}=vec{0}$



Vector product - function $times:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}^3$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})times vec{w} = alpha(vec{u}timesvec{w})+beta(vec{v}times vec{w})$



b) $vec{u} times vec{v} = -vec{v} times vec{u}$



c) $ (vec{u} times vec{v}) times vec{w} +(vec{v} times vec{w}) times vec{u}+(vec{w} times vec{u}) times vec{v}=vec{0}$



Question: We have endowed $mathbb{R}^3$ with these two structures (structure of scalar product and structure of vector product). I know that space endowed with dot product is called inner space, is there a name for a vector space endowed with vector product? Are vector product and scalar product (as I have defined them) connected structures? Can we somehow exspress vecotor prodcut via scalar product or slacar product via vector prodcut? Is there some formula which can reduce on product to another? What preciasly is connection between these two structures, and can that connection be generalized to $mathbb{R}^n$ (I do not know whether this question makes sense)?



Edit: I am hoping that if there is some connection between these two strcutures that answers not just say it, but rather explain why does this connection hold.



Thank you for any help.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe the concept of exterior algebra is the genralization you are looking for.
    $endgroup$
    – G.F
    5 hours ago










  • $begingroup$
    The properties listed for $times$ do hold, but they don't make for a satisfactory definition. After all, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ satisfies all three.
    $endgroup$
    – Travis
    5 hours ago










  • $begingroup$
    I think it's rather the Lie algebras.
    $endgroup$
    – Berci
    5 hours ago










  • $begingroup$
    Why has nobody mentioned the geometric product? That combines the dot and cross products, and it generalizes to any number of dimensions. The dot product is the symmetric part, and the cross product is (the dual of) the antisymmetric part, of the geometric product. $$acdot b=frac{ab+ba}{2}$$ $$atimes b=Ifrac{ab-ba}{2}=I(awedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra
    $endgroup$
    – mr_e_man
    1 hour ago












  • $begingroup$
    @mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it?
    $endgroup$
    – Thom
    1 hour ago














3












3








3


1



$begingroup$


I am intrested in relationship between scalar and vector product in $mathbb{R}^3$; I am going to give definitions which I will use in my question.



Scalar product - function $cdot:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})cdot vec{w} = alpha(vec{u}cdot vec{w})+beta(vec{v}cdot vec{w})$



b) $vec{u}cdotvec{v}=vec{v}cdotvec{u}$



c) $vec{u}cdot vec{u} geq 0$ and $vec{u}cdotvec{u}=0 iff vec{u}=vec{0}$



Vector product - function $times:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}^3$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})times vec{w} = alpha(vec{u}timesvec{w})+beta(vec{v}times vec{w})$



b) $vec{u} times vec{v} = -vec{v} times vec{u}$



c) $ (vec{u} times vec{v}) times vec{w} +(vec{v} times vec{w}) times vec{u}+(vec{w} times vec{u}) times vec{v}=vec{0}$



Question: We have endowed $mathbb{R}^3$ with these two structures (structure of scalar product and structure of vector product). I know that space endowed with dot product is called inner space, is there a name for a vector space endowed with vector product? Are vector product and scalar product (as I have defined them) connected structures? Can we somehow exspress vecotor prodcut via scalar product or slacar product via vector prodcut? Is there some formula which can reduce on product to another? What preciasly is connection between these two structures, and can that connection be generalized to $mathbb{R}^n$ (I do not know whether this question makes sense)?



Edit: I am hoping that if there is some connection between these two strcutures that answers not just say it, but rather explain why does this connection hold.



Thank you for any help.










share|cite|improve this question











$endgroup$




I am intrested in relationship between scalar and vector product in $mathbb{R}^3$; I am going to give definitions which I will use in my question.



Scalar product - function $cdot:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})cdot vec{w} = alpha(vec{u}cdot vec{w})+beta(vec{v}cdot vec{w})$



b) $vec{u}cdotvec{v}=vec{v}cdotvec{u}$



c) $vec{u}cdot vec{u} geq 0$ and $vec{u}cdotvec{u}=0 iff vec{u}=vec{0}$



Vector product - function $times:mathbb{R}^3 times mathbb{R}^3 to mathbb{R}^3$ which satisfay following properties:



a) $(alpha vec{u}+betavec{v})times vec{w} = alpha(vec{u}timesvec{w})+beta(vec{v}times vec{w})$



b) $vec{u} times vec{v} = -vec{v} times vec{u}$



c) $ (vec{u} times vec{v}) times vec{w} +(vec{v} times vec{w}) times vec{u}+(vec{w} times vec{u}) times vec{v}=vec{0}$



Question: We have endowed $mathbb{R}^3$ with these two structures (structure of scalar product and structure of vector product). I know that space endowed with dot product is called inner space, is there a name for a vector space endowed with vector product? Are vector product and scalar product (as I have defined them) connected structures? Can we somehow exspress vecotor prodcut via scalar product or slacar product via vector prodcut? Is there some formula which can reduce on product to another? What preciasly is connection between these two structures, and can that connection be generalized to $mathbb{R}^n$ (I do not know whether this question makes sense)?



Edit: I am hoping that if there is some connection between these two strcutures that answers not just say it, but rather explain why does this connection hold.



Thank you for any help.







linear-algebra vector-spaces inner-product-space






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share|cite|improve this question













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share|cite|improve this question








edited 5 hours ago







Thom

















asked 6 hours ago









ThomThom

32219




32219








  • 1




    $begingroup$
    Maybe the concept of exterior algebra is the genralization you are looking for.
    $endgroup$
    – G.F
    5 hours ago










  • $begingroup$
    The properties listed for $times$ do hold, but they don't make for a satisfactory definition. After all, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ satisfies all three.
    $endgroup$
    – Travis
    5 hours ago










  • $begingroup$
    I think it's rather the Lie algebras.
    $endgroup$
    – Berci
    5 hours ago










  • $begingroup$
    Why has nobody mentioned the geometric product? That combines the dot and cross products, and it generalizes to any number of dimensions. The dot product is the symmetric part, and the cross product is (the dual of) the antisymmetric part, of the geometric product. $$acdot b=frac{ab+ba}{2}$$ $$atimes b=Ifrac{ab-ba}{2}=I(awedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra
    $endgroup$
    – mr_e_man
    1 hour ago












  • $begingroup$
    @mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it?
    $endgroup$
    – Thom
    1 hour ago














  • 1




    $begingroup$
    Maybe the concept of exterior algebra is the genralization you are looking for.
    $endgroup$
    – G.F
    5 hours ago










  • $begingroup$
    The properties listed for $times$ do hold, but they don't make for a satisfactory definition. After all, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ satisfies all three.
    $endgroup$
    – Travis
    5 hours ago










  • $begingroup$
    I think it's rather the Lie algebras.
    $endgroup$
    – Berci
    5 hours ago










  • $begingroup$
    Why has nobody mentioned the geometric product? That combines the dot and cross products, and it generalizes to any number of dimensions. The dot product is the symmetric part, and the cross product is (the dual of) the antisymmetric part, of the geometric product. $$acdot b=frac{ab+ba}{2}$$ $$atimes b=Ifrac{ab-ba}{2}=I(awedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra
    $endgroup$
    – mr_e_man
    1 hour ago












  • $begingroup$
    @mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it?
    $endgroup$
    – Thom
    1 hour ago








1




1




$begingroup$
Maybe the concept of exterior algebra is the genralization you are looking for.
$endgroup$
– G.F
5 hours ago




$begingroup$
Maybe the concept of exterior algebra is the genralization you are looking for.
$endgroup$
– G.F
5 hours ago












$begingroup$
The properties listed for $times$ do hold, but they don't make for a satisfactory definition. After all, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ satisfies all three.
$endgroup$
– Travis
5 hours ago




$begingroup$
The properties listed for $times$ do hold, but they don't make for a satisfactory definition. After all, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ satisfies all three.
$endgroup$
– Travis
5 hours ago












$begingroup$
I think it's rather the Lie algebras.
$endgroup$
– Berci
5 hours ago




$begingroup$
I think it's rather the Lie algebras.
$endgroup$
– Berci
5 hours ago












$begingroup$
Why has nobody mentioned the geometric product? That combines the dot and cross products, and it generalizes to any number of dimensions. The dot product is the symmetric part, and the cross product is (the dual of) the antisymmetric part, of the geometric product. $$acdot b=frac{ab+ba}{2}$$ $$atimes b=Ifrac{ab-ba}{2}=I(awedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra
$endgroup$
– mr_e_man
1 hour ago






$begingroup$
Why has nobody mentioned the geometric product? That combines the dot and cross products, and it generalizes to any number of dimensions. The dot product is the symmetric part, and the cross product is (the dual of) the antisymmetric part, of the geometric product. $$acdot b=frac{ab+ba}{2}$$ $$atimes b=Ifrac{ab-ba}{2}=I(awedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra
$endgroup$
– mr_e_man
1 hour ago














$begingroup$
@mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it?
$endgroup$
– Thom
1 hour ago




$begingroup$
@mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it?
$endgroup$
– Thom
1 hour ago










4 Answers
4






active

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The usual inner (dot) and cross products on $Bbb R^3$ are defined by
$${bf x} cdot {bf y} = x_1 y_1 + x_2 y_2 + x_3 y_3, qquad
{bf x} times {bf y} = (x_2 y_3 - x_3 y_2, x_3 y_1 - x_1 y_3, x_1 y_2 - x_2 y_1) .$$

They are related by the properties
$$phantom{(ast)} qquad ({bf x} times {bf y}) cdot {bf x} = 0, qquad
({bf x} times {bf y}) cdot ({bf x} times {bf y}) = ({bf x} cdot {bf x})({bf y} cdot {bf y}) - ({bf x} cdot {bf y})^2 , qquad (ast)$$

and by the vector triple product identity,
$${bf x} times ({bf y} times {bf z}) = ({bf x} cdot {bf z}) {bf y} - ({bf x} cdot {bf y}) {bf z}.$$



By taking an appropriate trace one can derive from this formula one that recovers the dot product from the cross product:
$${bf x} cdot {bf y} = -frac{1}{2} operatorname{tr}({bf z} mapsto {bf x} times ({bf y} times {bf z})) .$$



One cannot, however, recover the cross product from the dot product (alone): If one replaces the "right-handed" cross product $times$ with the "left-handed" cross product ${bf x} times' {bf y} := -{bf x} times {bf y}$, we see that $times'$ still satisfies all of the above properties. Indeed, the first equation of $(ast)$ says that ${bf x} times {bf y}$ is a vector mutually orthogonal to ${bf x}$ and ${bf y}$, and the second specifies its length, but these two conditions determine ${bf x} times {bf y}$ only up to sign. Thus, we can recover $times$ from $cdot$ using $(ast)$ and a choice of orientation, i.e., a choice of "handedness". (If you know a little group theory, this situation can be viewed another way: The group of linear transformations that preserve the cross product is $SO(3)$, but the group of those that preserve the dot product is $O(3)$, in which $SO(3)$ is an index-$2$ subgroup.)



The structure defined by the three given conditions (a)-(c) satisfied by the cross product is called a Lie algebra, and usually one writes the product of $a, b$ in a Lie algebra as $[a, b]$. These are not in general called cross products however, and most don't have a nice relationship with a particular inner product like $times$ does. (Indeed, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ defines the abelian Lie algebra on $Bbb R^3$.) Instead, the term cross product is usually reserved for a map $Bbb R^n times cdots times Bbb R^n to Bbb R^n$ that satisfies $(ast)$, where we replace the second equation in $(ast)$ with an appropriate generalization.



To give an interesting example, suppose with identify $Bbb R^3$ with the vector space of tracefree $2 times 2$ real matrices, that is, those of the form $$pmatrix{a&b\c&-a} .$$ We can define a "cross product" on this set by
$$A times B = operatorname{tf}(AB) = A B - tfrac{1}{2} operatorname{tr}(AB) I$$---here $operatorname{tf} C$ just denotes the tracefree part of C. Using the above trace formula gives a "dot product" $A cdot B = tfrac{1}{2} operatorname{tr}(AB)$. Here I put "dot product" in quotation marks because this product is not positive-definite, i.e., doesn't satisfy property (c), but it satisfies the nondegeneracy condition that if ${bf x} cdot {bf y} = 0$ for all ${bf y}$ then ${bf x} = 0$.






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  • $begingroup$
    Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
    $endgroup$
    – Thom
    2 hours ago








  • 1




    $begingroup$
    No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
    $endgroup$
    – Travis
    1 hour ago










  • $begingroup$
    Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
    $endgroup$
    – Thom
    1 hour ago






  • 1




    $begingroup$
    No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
    $endgroup$
    – Travis
    1 hour ago






  • 1




    $begingroup$
    You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
    $endgroup$
    – Travis
    1 hour ago



















3












$begingroup$

You can easily define a scalar product in $mathbb{R}^n$, for each $ninmathbb N$, be the vector product is specific to $mathbb{R}^3$. There is a generalization to $mathbb{R}^n$, but that's a map from $(mathbb{R}^n)^{n-1}$ into $mathbb{R}^n$.



A connection betwen both operations (assuming that you are dealing with the usual dot product here) is given by the triple product: given $v,w,uinmathbb{R}^3$, $bigllvert v.(wtimes u)bigrrvert$ is the volume of the parallelepiped defined by them.






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    1












    $begingroup$

    If you want to relate the two products, note that



    $$|atimes b|^2 + (acdot b)^2 = |a|^2|b|^2.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Well, why does that hold?
      $endgroup$
      – Thom
      5 hours ago










    • $begingroup$
      @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
      $endgroup$
      – B. Goddard
      5 hours ago










    • $begingroup$
      Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
      $endgroup$
      – Thom
      5 hours ago










    • $begingroup$
      @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
      $endgroup$
      – B. Goddard
      5 hours ago










    • $begingroup$
      we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
      $endgroup$
      – Thom
      5 hours ago





















    1












    $begingroup$

    We say $Bbb R^3$ is an exterior algebra under $times$; in such terminology it's the wedge product $land$.



    Scalar-vector connections include the scalar triple product $acdot btimes c$ being fully antisymmetric, $(acdot b)^2+(atimes b)cdot(atimes b)=(acdot a)(bcdot b)$ and $atimes (btimes c)=(acdot c)b-(acdot b)c$.



    Proofs, with implicit summation over repeated indices:





    • $acdot (btimes c)=varepsilon_{ijk}a_ib_jc_k$, then use $varepsilon_{ijk}$ being fully antisymmetric


    • $(acdot b)^2+(atimes b)cdot(atimes b)=a_jb_ka_mb_n(delta_{jk}delta_{lm}+varepsilon_{ijk}varepsilon_{ilm})$, the bracketed coefficient famously being $delta_{jm}delta_{kn}$

    • The left-hand side's $i$th component is $varepsilon_{ijk}varepsilon_{klm}a_jb_mc_n=(delta_{il}delta_{jm}-delta_{im}delta_{jl})a_jb_mc_n$ as required






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do we prove these connections and are those the only one?
      $endgroup$
      – Thom
      5 hours ago










    • $begingroup$
      @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
      $endgroup$
      – J.G.
      5 hours ago










    • $begingroup$
      How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
      $endgroup$
      – Thom
      5 hours ago










    • $begingroup$
      @Thom That's a fair request. See my edit.
      $endgroup$
      – J.G.
      5 hours ago











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    4 Answers
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    4 Answers
    4






    active

    oldest

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    The usual inner (dot) and cross products on $Bbb R^3$ are defined by
    $${bf x} cdot {bf y} = x_1 y_1 + x_2 y_2 + x_3 y_3, qquad
    {bf x} times {bf y} = (x_2 y_3 - x_3 y_2, x_3 y_1 - x_1 y_3, x_1 y_2 - x_2 y_1) .$$

    They are related by the properties
    $$phantom{(ast)} qquad ({bf x} times {bf y}) cdot {bf x} = 0, qquad
    ({bf x} times {bf y}) cdot ({bf x} times {bf y}) = ({bf x} cdot {bf x})({bf y} cdot {bf y}) - ({bf x} cdot {bf y})^2 , qquad (ast)$$

    and by the vector triple product identity,
    $${bf x} times ({bf y} times {bf z}) = ({bf x} cdot {bf z}) {bf y} - ({bf x} cdot {bf y}) {bf z}.$$



    By taking an appropriate trace one can derive from this formula one that recovers the dot product from the cross product:
    $${bf x} cdot {bf y} = -frac{1}{2} operatorname{tr}({bf z} mapsto {bf x} times ({bf y} times {bf z})) .$$



    One cannot, however, recover the cross product from the dot product (alone): If one replaces the "right-handed" cross product $times$ with the "left-handed" cross product ${bf x} times' {bf y} := -{bf x} times {bf y}$, we see that $times'$ still satisfies all of the above properties. Indeed, the first equation of $(ast)$ says that ${bf x} times {bf y}$ is a vector mutually orthogonal to ${bf x}$ and ${bf y}$, and the second specifies its length, but these two conditions determine ${bf x} times {bf y}$ only up to sign. Thus, we can recover $times$ from $cdot$ using $(ast)$ and a choice of orientation, i.e., a choice of "handedness". (If you know a little group theory, this situation can be viewed another way: The group of linear transformations that preserve the cross product is $SO(3)$, but the group of those that preserve the dot product is $O(3)$, in which $SO(3)$ is an index-$2$ subgroup.)



    The structure defined by the three given conditions (a)-(c) satisfied by the cross product is called a Lie algebra, and usually one writes the product of $a, b$ in a Lie algebra as $[a, b]$. These are not in general called cross products however, and most don't have a nice relationship with a particular inner product like $times$ does. (Indeed, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ defines the abelian Lie algebra on $Bbb R^3$.) Instead, the term cross product is usually reserved for a map $Bbb R^n times cdots times Bbb R^n to Bbb R^n$ that satisfies $(ast)$, where we replace the second equation in $(ast)$ with an appropriate generalization.



    To give an interesting example, suppose with identify $Bbb R^3$ with the vector space of tracefree $2 times 2$ real matrices, that is, those of the form $$pmatrix{a&b\c&-a} .$$ We can define a "cross product" on this set by
    $$A times B = operatorname{tf}(AB) = A B - tfrac{1}{2} operatorname{tr}(AB) I$$---here $operatorname{tf} C$ just denotes the tracefree part of C. Using the above trace formula gives a "dot product" $A cdot B = tfrac{1}{2} operatorname{tr}(AB)$. Here I put "dot product" in quotation marks because this product is not positive-definite, i.e., doesn't satisfy property (c), but it satisfies the nondegeneracy condition that if ${bf x} cdot {bf y} = 0$ for all ${bf y}$ then ${bf x} = 0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
      $endgroup$
      – Thom
      2 hours ago








    • 1




      $begingroup$
      No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
      $endgroup$
      – Travis
      1 hour ago










    • $begingroup$
      Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
      $endgroup$
      – Thom
      1 hour ago






    • 1




      $begingroup$
      No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
      $endgroup$
      – Travis
      1 hour ago






    • 1




      $begingroup$
      You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
      $endgroup$
      – Travis
      1 hour ago
















    1












    $begingroup$

    The usual inner (dot) and cross products on $Bbb R^3$ are defined by
    $${bf x} cdot {bf y} = x_1 y_1 + x_2 y_2 + x_3 y_3, qquad
    {bf x} times {bf y} = (x_2 y_3 - x_3 y_2, x_3 y_1 - x_1 y_3, x_1 y_2 - x_2 y_1) .$$

    They are related by the properties
    $$phantom{(ast)} qquad ({bf x} times {bf y}) cdot {bf x} = 0, qquad
    ({bf x} times {bf y}) cdot ({bf x} times {bf y}) = ({bf x} cdot {bf x})({bf y} cdot {bf y}) - ({bf x} cdot {bf y})^2 , qquad (ast)$$

    and by the vector triple product identity,
    $${bf x} times ({bf y} times {bf z}) = ({bf x} cdot {bf z}) {bf y} - ({bf x} cdot {bf y}) {bf z}.$$



    By taking an appropriate trace one can derive from this formula one that recovers the dot product from the cross product:
    $${bf x} cdot {bf y} = -frac{1}{2} operatorname{tr}({bf z} mapsto {bf x} times ({bf y} times {bf z})) .$$



    One cannot, however, recover the cross product from the dot product (alone): If one replaces the "right-handed" cross product $times$ with the "left-handed" cross product ${bf x} times' {bf y} := -{bf x} times {bf y}$, we see that $times'$ still satisfies all of the above properties. Indeed, the first equation of $(ast)$ says that ${bf x} times {bf y}$ is a vector mutually orthogonal to ${bf x}$ and ${bf y}$, and the second specifies its length, but these two conditions determine ${bf x} times {bf y}$ only up to sign. Thus, we can recover $times$ from $cdot$ using $(ast)$ and a choice of orientation, i.e., a choice of "handedness". (If you know a little group theory, this situation can be viewed another way: The group of linear transformations that preserve the cross product is $SO(3)$, but the group of those that preserve the dot product is $O(3)$, in which $SO(3)$ is an index-$2$ subgroup.)



    The structure defined by the three given conditions (a)-(c) satisfied by the cross product is called a Lie algebra, and usually one writes the product of $a, b$ in a Lie algebra as $[a, b]$. These are not in general called cross products however, and most don't have a nice relationship with a particular inner product like $times$ does. (Indeed, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ defines the abelian Lie algebra on $Bbb R^3$.) Instead, the term cross product is usually reserved for a map $Bbb R^n times cdots times Bbb R^n to Bbb R^n$ that satisfies $(ast)$, where we replace the second equation in $(ast)$ with an appropriate generalization.



    To give an interesting example, suppose with identify $Bbb R^3$ with the vector space of tracefree $2 times 2$ real matrices, that is, those of the form $$pmatrix{a&b\c&-a} .$$ We can define a "cross product" on this set by
    $$A times B = operatorname{tf}(AB) = A B - tfrac{1}{2} operatorname{tr}(AB) I$$---here $operatorname{tf} C$ just denotes the tracefree part of C. Using the above trace formula gives a "dot product" $A cdot B = tfrac{1}{2} operatorname{tr}(AB)$. Here I put "dot product" in quotation marks because this product is not positive-definite, i.e., doesn't satisfy property (c), but it satisfies the nondegeneracy condition that if ${bf x} cdot {bf y} = 0$ for all ${bf y}$ then ${bf x} = 0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
      $endgroup$
      – Thom
      2 hours ago








    • 1




      $begingroup$
      No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
      $endgroup$
      – Travis
      1 hour ago










    • $begingroup$
      Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
      $endgroup$
      – Thom
      1 hour ago






    • 1




      $begingroup$
      No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
      $endgroup$
      – Travis
      1 hour ago






    • 1




      $begingroup$
      You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
      $endgroup$
      – Travis
      1 hour ago














    1












    1








    1





    $begingroup$

    The usual inner (dot) and cross products on $Bbb R^3$ are defined by
    $${bf x} cdot {bf y} = x_1 y_1 + x_2 y_2 + x_3 y_3, qquad
    {bf x} times {bf y} = (x_2 y_3 - x_3 y_2, x_3 y_1 - x_1 y_3, x_1 y_2 - x_2 y_1) .$$

    They are related by the properties
    $$phantom{(ast)} qquad ({bf x} times {bf y}) cdot {bf x} = 0, qquad
    ({bf x} times {bf y}) cdot ({bf x} times {bf y}) = ({bf x} cdot {bf x})({bf y} cdot {bf y}) - ({bf x} cdot {bf y})^2 , qquad (ast)$$

    and by the vector triple product identity,
    $${bf x} times ({bf y} times {bf z}) = ({bf x} cdot {bf z}) {bf y} - ({bf x} cdot {bf y}) {bf z}.$$



    By taking an appropriate trace one can derive from this formula one that recovers the dot product from the cross product:
    $${bf x} cdot {bf y} = -frac{1}{2} operatorname{tr}({bf z} mapsto {bf x} times ({bf y} times {bf z})) .$$



    One cannot, however, recover the cross product from the dot product (alone): If one replaces the "right-handed" cross product $times$ with the "left-handed" cross product ${bf x} times' {bf y} := -{bf x} times {bf y}$, we see that $times'$ still satisfies all of the above properties. Indeed, the first equation of $(ast)$ says that ${bf x} times {bf y}$ is a vector mutually orthogonal to ${bf x}$ and ${bf y}$, and the second specifies its length, but these two conditions determine ${bf x} times {bf y}$ only up to sign. Thus, we can recover $times$ from $cdot$ using $(ast)$ and a choice of orientation, i.e., a choice of "handedness". (If you know a little group theory, this situation can be viewed another way: The group of linear transformations that preserve the cross product is $SO(3)$, but the group of those that preserve the dot product is $O(3)$, in which $SO(3)$ is an index-$2$ subgroup.)



    The structure defined by the three given conditions (a)-(c) satisfied by the cross product is called a Lie algebra, and usually one writes the product of $a, b$ in a Lie algebra as $[a, b]$. These are not in general called cross products however, and most don't have a nice relationship with a particular inner product like $times$ does. (Indeed, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ defines the abelian Lie algebra on $Bbb R^3$.) Instead, the term cross product is usually reserved for a map $Bbb R^n times cdots times Bbb R^n to Bbb R^n$ that satisfies $(ast)$, where we replace the second equation in $(ast)$ with an appropriate generalization.



    To give an interesting example, suppose with identify $Bbb R^3$ with the vector space of tracefree $2 times 2$ real matrices, that is, those of the form $$pmatrix{a&b\c&-a} .$$ We can define a "cross product" on this set by
    $$A times B = operatorname{tf}(AB) = A B - tfrac{1}{2} operatorname{tr}(AB) I$$---here $operatorname{tf} C$ just denotes the tracefree part of C. Using the above trace formula gives a "dot product" $A cdot B = tfrac{1}{2} operatorname{tr}(AB)$. Here I put "dot product" in quotation marks because this product is not positive-definite, i.e., doesn't satisfy property (c), but it satisfies the nondegeneracy condition that if ${bf x} cdot {bf y} = 0$ for all ${bf y}$ then ${bf x} = 0$.






    share|cite|improve this answer









    $endgroup$



    The usual inner (dot) and cross products on $Bbb R^3$ are defined by
    $${bf x} cdot {bf y} = x_1 y_1 + x_2 y_2 + x_3 y_3, qquad
    {bf x} times {bf y} = (x_2 y_3 - x_3 y_2, x_3 y_1 - x_1 y_3, x_1 y_2 - x_2 y_1) .$$

    They are related by the properties
    $$phantom{(ast)} qquad ({bf x} times {bf y}) cdot {bf x} = 0, qquad
    ({bf x} times {bf y}) cdot ({bf x} times {bf y}) = ({bf x} cdot {bf x})({bf y} cdot {bf y}) - ({bf x} cdot {bf y})^2 , qquad (ast)$$

    and by the vector triple product identity,
    $${bf x} times ({bf y} times {bf z}) = ({bf x} cdot {bf z}) {bf y} - ({bf x} cdot {bf y}) {bf z}.$$



    By taking an appropriate trace one can derive from this formula one that recovers the dot product from the cross product:
    $${bf x} cdot {bf y} = -frac{1}{2} operatorname{tr}({bf z} mapsto {bf x} times ({bf y} times {bf z})) .$$



    One cannot, however, recover the cross product from the dot product (alone): If one replaces the "right-handed" cross product $times$ with the "left-handed" cross product ${bf x} times' {bf y} := -{bf x} times {bf y}$, we see that $times'$ still satisfies all of the above properties. Indeed, the first equation of $(ast)$ says that ${bf x} times {bf y}$ is a vector mutually orthogonal to ${bf x}$ and ${bf y}$, and the second specifies its length, but these two conditions determine ${bf x} times {bf y}$ only up to sign. Thus, we can recover $times$ from $cdot$ using $(ast)$ and a choice of orientation, i.e., a choice of "handedness". (If you know a little group theory, this situation can be viewed another way: The group of linear transformations that preserve the cross product is $SO(3)$, but the group of those that preserve the dot product is $O(3)$, in which $SO(3)$ is an index-$2$ subgroup.)



    The structure defined by the three given conditions (a)-(c) satisfied by the cross product is called a Lie algebra, and usually one writes the product of $a, b$ in a Lie algebra as $[a, b]$. These are not in general called cross products however, and most don't have a nice relationship with a particular inner product like $times$ does. (Indeed, the zero map $Bbb R^3 times Bbb R^3 to Bbb R^3$ defines the abelian Lie algebra on $Bbb R^3$.) Instead, the term cross product is usually reserved for a map $Bbb R^n times cdots times Bbb R^n to Bbb R^n$ that satisfies $(ast)$, where we replace the second equation in $(ast)$ with an appropriate generalization.



    To give an interesting example, suppose with identify $Bbb R^3$ with the vector space of tracefree $2 times 2$ real matrices, that is, those of the form $$pmatrix{a&b\c&-a} .$$ We can define a "cross product" on this set by
    $$A times B = operatorname{tf}(AB) = A B - tfrac{1}{2} operatorname{tr}(AB) I$$---here $operatorname{tf} C$ just denotes the tracefree part of C. Using the above trace formula gives a "dot product" $A cdot B = tfrac{1}{2} operatorname{tr}(AB)$. Here I put "dot product" in quotation marks because this product is not positive-definite, i.e., doesn't satisfy property (c), but it satisfies the nondegeneracy condition that if ${bf x} cdot {bf y} = 0$ for all ${bf y}$ then ${bf x} = 0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 4 hours ago









    TravisTravis

    60k767146




    60k767146












    • $begingroup$
      Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
      $endgroup$
      – Thom
      2 hours ago








    • 1




      $begingroup$
      No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
      $endgroup$
      – Travis
      1 hour ago










    • $begingroup$
      Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
      $endgroup$
      – Thom
      1 hour ago






    • 1




      $begingroup$
      No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
      $endgroup$
      – Travis
      1 hour ago






    • 1




      $begingroup$
      You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
      $endgroup$
      – Travis
      1 hour ago


















    • $begingroup$
      Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
      $endgroup$
      – Thom
      2 hours ago








    • 1




      $begingroup$
      No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
      $endgroup$
      – Travis
      1 hour ago










    • $begingroup$
      Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
      $endgroup$
      – Thom
      1 hour ago






    • 1




      $begingroup$
      No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
      $endgroup$
      – Travis
      1 hour ago






    • 1




      $begingroup$
      You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
      $endgroup$
      – Travis
      1 hour ago
















    $begingroup$
    Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
    $endgroup$
    – Thom
    2 hours ago






    $begingroup$
    Fantastic answer. However I still have some questions. Once we choose usal inner (dot) and cross product we can easily prove two formulas in * and triple product identity and formula with trace which recovers dot product from cross product. But can we prove these formulas without having to choose usal dot and cross product, only via using definig properties which I listed in the answer (if yes, how)?
    $endgroup$
    – Thom
    2 hours ago






    1




    1




    $begingroup$
    No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
    $endgroup$
    – Travis
    1 hour ago




    $begingroup$
    No, one cannot prove these simply because they don't hold for all choices of $cdot$ and $times$ satisfying the properties you listed. For example, the zero map satisfies the three axioms you give for $times$, but the zero map satisfies neither the second property in $(ast)$ nor the triple product identity.
    $endgroup$
    – Travis
    1 hour ago












    $begingroup$
    Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
    $endgroup$
    – Thom
    1 hour ago




    $begingroup$
    Is there any general connection, or relation between dot product and cross product which must, nessecarily work for every choice of dot and cross product or they can be choosen complatley independently?
    $endgroup$
    – Thom
    1 hour ago




    1




    1




    $begingroup$
    No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
    $endgroup$
    – Travis
    1 hour ago




    $begingroup$
    No, you cannot pick $times$ and $cdot$ independently, at least if you want the usual identities to be satisfied: Even if you pick a cross product $times$ for which the above trace formula gives an inner product (there are infinitely many choices, but they're all qualitatively equivalent), you're obliged to take that inner product if you want $(ast)$ and the triple product identity to hold. Once you do, though, all of the usual identities (in particular those in my answer) follow.
    $endgroup$
    – Travis
    1 hour ago




    1




    1




    $begingroup$
    You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
    $endgroup$
    – Travis
    1 hour ago




    $begingroup$
    You're welcome, I'm glad you found it helpful. You might be interested, by the way, in reading up on how the dot and cross product on $Bbb R^3$ can be described in terms of quaternions. From that point of view, all of the properties of those products, as well as the identities relating them, can be viewed as easy consequences of quaternion identities.
    $endgroup$
    – Travis
    1 hour ago











    3












    $begingroup$

    You can easily define a scalar product in $mathbb{R}^n$, for each $ninmathbb N$, be the vector product is specific to $mathbb{R}^3$. There is a generalization to $mathbb{R}^n$, but that's a map from $(mathbb{R}^n)^{n-1}$ into $mathbb{R}^n$.



    A connection betwen both operations (assuming that you are dealing with the usual dot product here) is given by the triple product: given $v,w,uinmathbb{R}^3$, $bigllvert v.(wtimes u)bigrrvert$ is the volume of the parallelepiped defined by them.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You can easily define a scalar product in $mathbb{R}^n$, for each $ninmathbb N$, be the vector product is specific to $mathbb{R}^3$. There is a generalization to $mathbb{R}^n$, but that's a map from $(mathbb{R}^n)^{n-1}$ into $mathbb{R}^n$.



      A connection betwen both operations (assuming that you are dealing with the usual dot product here) is given by the triple product: given $v,w,uinmathbb{R}^3$, $bigllvert v.(wtimes u)bigrrvert$ is the volume of the parallelepiped defined by them.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You can easily define a scalar product in $mathbb{R}^n$, for each $ninmathbb N$, be the vector product is specific to $mathbb{R}^3$. There is a generalization to $mathbb{R}^n$, but that's a map from $(mathbb{R}^n)^{n-1}$ into $mathbb{R}^n$.



        A connection betwen both operations (assuming that you are dealing with the usual dot product here) is given by the triple product: given $v,w,uinmathbb{R}^3$, $bigllvert v.(wtimes u)bigrrvert$ is the volume of the parallelepiped defined by them.






        share|cite|improve this answer









        $endgroup$



        You can easily define a scalar product in $mathbb{R}^n$, for each $ninmathbb N$, be the vector product is specific to $mathbb{R}^3$. There is a generalization to $mathbb{R}^n$, but that's a map from $(mathbb{R}^n)^{n-1}$ into $mathbb{R}^n$.



        A connection betwen both operations (assuming that you are dealing with the usual dot product here) is given by the triple product: given $v,w,uinmathbb{R}^3$, $bigllvert v.(wtimes u)bigrrvert$ is the volume of the parallelepiped defined by them.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        José Carlos SantosJosé Carlos Santos

        155k22124227




        155k22124227























            1












            $begingroup$

            If you want to relate the two products, note that



            $$|atimes b|^2 + (acdot b)^2 = |a|^2|b|^2.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Well, why does that hold?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
              $endgroup$
              – Thom
              5 hours ago


















            1












            $begingroup$

            If you want to relate the two products, note that



            $$|atimes b|^2 + (acdot b)^2 = |a|^2|b|^2.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Well, why does that hold?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
              $endgroup$
              – Thom
              5 hours ago
















            1












            1








            1





            $begingroup$

            If you want to relate the two products, note that



            $$|atimes b|^2 + (acdot b)^2 = |a|^2|b|^2.$$






            share|cite|improve this answer









            $endgroup$



            If you want to relate the two products, note that



            $$|atimes b|^2 + (acdot b)^2 = |a|^2|b|^2.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 5 hours ago









            B. GoddardB. Goddard

            18.6k21340




            18.6k21340












            • $begingroup$
              Well, why does that hold?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
              $endgroup$
              – Thom
              5 hours ago




















            • $begingroup$
              Well, why does that hold?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
              $endgroup$
              – B. Goddard
              5 hours ago










            • $begingroup$
              we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
              $endgroup$
              – Thom
              5 hours ago


















            $begingroup$
            Well, why does that hold?
            $endgroup$
            – Thom
            5 hours ago




            $begingroup$
            Well, why does that hold?
            $endgroup$
            – Thom
            5 hours ago












            $begingroup$
            @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
            $endgroup$
            – B. Goddard
            5 hours ago




            $begingroup$
            @Thom The left side equals $(|a||b|sintheta)^2 + (|a||b|costheta)^2.$
            $endgroup$
            – B. Goddard
            5 hours ago












            $begingroup$
            Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
            $endgroup$
            – Thom
            5 hours ago




            $begingroup$
            Where did I assume standard definition of scalar product? Or somehow for all scalar products we can write $vec{a}cdot vec{b}=|a||b|cos(theta)$ ?
            $endgroup$
            – Thom
            5 hours ago












            $begingroup$
            @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
            $endgroup$
            – B. Goddard
            5 hours ago




            $begingroup$
            @Thom You didn't. But now you're assuming a certain definition of the angle between two vectors. Your products imply some sort of orthagonality.
            $endgroup$
            – B. Goddard
            5 hours ago












            $begingroup$
            we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
            $endgroup$
            – Thom
            5 hours ago






            $begingroup$
            we can use this: $cos(theta) = (vec{a} cdot vec{b})/(|a||b|)$ to define something which we call angle $theta$ between two vectors. Why would that same $theta$ go nesscarily into definition of vector product in this manner $vec{a} times vec{b} = |a||b|sin(theta)$?
            $endgroup$
            – Thom
            5 hours ago













            1












            $begingroup$

            We say $Bbb R^3$ is an exterior algebra under $times$; in such terminology it's the wedge product $land$.



            Scalar-vector connections include the scalar triple product $acdot btimes c$ being fully antisymmetric, $(acdot b)^2+(atimes b)cdot(atimes b)=(acdot a)(bcdot b)$ and $atimes (btimes c)=(acdot c)b-(acdot b)c$.



            Proofs, with implicit summation over repeated indices:





            • $acdot (btimes c)=varepsilon_{ijk}a_ib_jc_k$, then use $varepsilon_{ijk}$ being fully antisymmetric


            • $(acdot b)^2+(atimes b)cdot(atimes b)=a_jb_ka_mb_n(delta_{jk}delta_{lm}+varepsilon_{ijk}varepsilon_{ilm})$, the bracketed coefficient famously being $delta_{jm}delta_{kn}$

            • The left-hand side's $i$th component is $varepsilon_{ijk}varepsilon_{klm}a_jb_mc_n=(delta_{il}delta_{jm}-delta_{im}delta_{jl})a_jb_mc_n$ as required






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do we prove these connections and are those the only one?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
              $endgroup$
              – J.G.
              5 hours ago










            • $begingroup$
              How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom That's a fair request. See my edit.
              $endgroup$
              – J.G.
              5 hours ago
















            1












            $begingroup$

            We say $Bbb R^3$ is an exterior algebra under $times$; in such terminology it's the wedge product $land$.



            Scalar-vector connections include the scalar triple product $acdot btimes c$ being fully antisymmetric, $(acdot b)^2+(atimes b)cdot(atimes b)=(acdot a)(bcdot b)$ and $atimes (btimes c)=(acdot c)b-(acdot b)c$.



            Proofs, with implicit summation over repeated indices:





            • $acdot (btimes c)=varepsilon_{ijk}a_ib_jc_k$, then use $varepsilon_{ijk}$ being fully antisymmetric


            • $(acdot b)^2+(atimes b)cdot(atimes b)=a_jb_ka_mb_n(delta_{jk}delta_{lm}+varepsilon_{ijk}varepsilon_{ilm})$, the bracketed coefficient famously being $delta_{jm}delta_{kn}$

            • The left-hand side's $i$th component is $varepsilon_{ijk}varepsilon_{klm}a_jb_mc_n=(delta_{il}delta_{jm}-delta_{im}delta_{jl})a_jb_mc_n$ as required






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do we prove these connections and are those the only one?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
              $endgroup$
              – J.G.
              5 hours ago










            • $begingroup$
              How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom That's a fair request. See my edit.
              $endgroup$
              – J.G.
              5 hours ago














            1












            1








            1





            $begingroup$

            We say $Bbb R^3$ is an exterior algebra under $times$; in such terminology it's the wedge product $land$.



            Scalar-vector connections include the scalar triple product $acdot btimes c$ being fully antisymmetric, $(acdot b)^2+(atimes b)cdot(atimes b)=(acdot a)(bcdot b)$ and $atimes (btimes c)=(acdot c)b-(acdot b)c$.



            Proofs, with implicit summation over repeated indices:





            • $acdot (btimes c)=varepsilon_{ijk}a_ib_jc_k$, then use $varepsilon_{ijk}$ being fully antisymmetric


            • $(acdot b)^2+(atimes b)cdot(atimes b)=a_jb_ka_mb_n(delta_{jk}delta_{lm}+varepsilon_{ijk}varepsilon_{ilm})$, the bracketed coefficient famously being $delta_{jm}delta_{kn}$

            • The left-hand side's $i$th component is $varepsilon_{ijk}varepsilon_{klm}a_jb_mc_n=(delta_{il}delta_{jm}-delta_{im}delta_{jl})a_jb_mc_n$ as required






            share|cite|improve this answer











            $endgroup$



            We say $Bbb R^3$ is an exterior algebra under $times$; in such terminology it's the wedge product $land$.



            Scalar-vector connections include the scalar triple product $acdot btimes c$ being fully antisymmetric, $(acdot b)^2+(atimes b)cdot(atimes b)=(acdot a)(bcdot b)$ and $atimes (btimes c)=(acdot c)b-(acdot b)c$.



            Proofs, with implicit summation over repeated indices:





            • $acdot (btimes c)=varepsilon_{ijk}a_ib_jc_k$, then use $varepsilon_{ijk}$ being fully antisymmetric


            • $(acdot b)^2+(atimes b)cdot(atimes b)=a_jb_ka_mb_n(delta_{jk}delta_{lm}+varepsilon_{ijk}varepsilon_{ilm})$, the bracketed coefficient famously being $delta_{jm}delta_{kn}$

            • The left-hand side's $i$th component is $varepsilon_{ijk}varepsilon_{klm}a_jb_mc_n=(delta_{il}delta_{jm}-delta_{im}delta_{jl})a_jb_mc_n$ as required







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 5 hours ago

























            answered 5 hours ago









            J.G.J.G.

            24.4k22539




            24.4k22539












            • $begingroup$
              How do we prove these connections and are those the only one?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
              $endgroup$
              – J.G.
              5 hours ago










            • $begingroup$
              How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom That's a fair request. See my edit.
              $endgroup$
              – J.G.
              5 hours ago


















            • $begingroup$
              How do we prove these connections and are those the only one?
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
              $endgroup$
              – J.G.
              5 hours ago










            • $begingroup$
              How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
              $endgroup$
              – Thom
              5 hours ago










            • $begingroup$
              @Thom That's a fair request. See my edit.
              $endgroup$
              – J.G.
              5 hours ago
















            $begingroup$
            How do we prove these connections and are those the only one?
            $endgroup$
            – Thom
            5 hours ago




            $begingroup$
            How do we prove these connections and are those the only one?
            $endgroup$
            – Thom
            5 hours ago












            $begingroup$
            @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
            $endgroup$
            – J.G.
            5 hours ago




            $begingroup$
            @Thorn They can all be proven from Levi-CIvita identities, but I don't think there are further connections that don't reduce to these.
            $endgroup$
            – J.G.
            5 hours ago












            $begingroup$
            How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
            $endgroup$
            – Thom
            5 hours ago




            $begingroup$
            How can they be proven? I was expacting that if there were some connection between that answers would not just say it but tell why it holds.
            $endgroup$
            – Thom
            5 hours ago












            $begingroup$
            @Thom That's a fair request. See my edit.
            $endgroup$
            – J.G.
            5 hours ago




            $begingroup$
            @Thom That's a fair request. See my edit.
            $endgroup$
            – J.G.
            5 hours ago


















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