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

b) $\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}$

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

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

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

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.

  • 1
    $\begingroup$ Maybe the concept of exterior algebra is the genralization you are looking for. $\endgroup$ – J.F Jan 19 '19 at 17:38
  • $\begingroup$ 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 Willse Jan 19 '19 at 17:51
  • $\begingroup$ I think it's rather the Lie algebras. $\endgroup$ – Berci Jan 19 '19 at 17:52
  • $\begingroup$ 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. $$a\cdot b=\frac{ab+ba}{2}$$ $$a\times b=I\frac{ab-ba}{2}=I(a\wedge b)$$ See en.wikipedia.org/wiki/Geometric_algebra $\endgroup$ – mr_e_man Jan 19 '19 at 22:06
  • $\begingroup$ @mr_e_man What is geometric product (and why it is called that) ? Where can I learn about it? $\endgroup$ – Thom Jan 19 '19 at 22:08

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 ${\bf x} = 0$.

| cite | improve this answer | |
  • $\begingroup$ 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 Jan 19 '19 at 21:21
  • 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 Willse Jan 19 '19 at 21:40
  • $\begingroup$ 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 Jan 19 '19 at 21:45
  • 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 Willse Jan 19 '19 at 21:47
  • 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 Willse Jan 19 '19 at 21:59

You can easily define a scalar product in $\mathbb{R}^n$, for each $n\in\mathbb N$, 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 $v,w,u\in\mathbb{R}^3$, $\bigl\lvert v.(w\times u)\bigr\rvert$ is the volume of the parallelepiped defined by them.

| cite | improve this answer | |

If you want to relate the two products, note that

$$|a\times b|^2 + (a\cdot b)^2 = |a|^2|b|^2.$$

| cite | improve this answer | |
  • $\begingroup$ Well, why does that hold? $\endgroup$ – Thom Jan 19 '19 at 17:57
  • $\begingroup$ @Thom The left side equals $(|a||b|\sin\theta)^2 + (|a||b|\cos\theta)^2.$ $\endgroup$ – B. Goddard Jan 19 '19 at 17:59
  • $\begingroup$ 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 Jan 19 '19 at 18:01
  • $\begingroup$ @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 Jan 19 '19 at 18:07
  • $\begingroup$ 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 Jan 19 '19 at 18:15

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 $a\cdot b\times c$ being fully antisymmetric, $(a\cdot b)^2+(a\times b)\cdot(a\times b)=(a\cdot a)(b\cdot b)$ and $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Proofs, with implicit summation over repeated indices:

  • $a\cdot (b\times c)=\varepsilon_{ijk}a_ib_jc_k$, then use $\varepsilon_{ijk}$ being fully antisymmetric
  • $(a\cdot b)^2+(a\times b)\cdot(a\times b)=a_jb_ka_mb_n(\delta_{jk}\delta_{lm}+\varepsilon_{ijk}\varepsilon_{ilm})$, 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
| cite | improve this answer | |
  • $\begingroup$ How do we prove these connections and are those the only one? $\endgroup$ – Thom Jan 19 '19 at 17:46
  • $\begingroup$ @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. Jan 19 '19 at 17:50
  • $\begingroup$ @Thom That's a fair request. See my edit. $\endgroup$ – J.G. Jan 19 '19 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.