Assume that $A$, $B$, $C$, and $D$ are four vectors in $\mathbb{R}^3$. Then I want to show that

$$ {\bf{M}} \equiv (A \times B) \otimes C + (B \times C) \otimes A + (C \times A) \otimes B = [(A \times B) \cdot C] {\bf{I}} \tag{1}$$

where $\otimes$ is the tensor product, $\cdot$ is the scalar product and $\times$ is the cross product. Also, ${\bf{I}}$ is the second order identity tensor.


The motivation behind this question is the vector identity

$$ [(A \times B) \cdot C] D = (A \times B) (C \cdot D) + (B \times C) (A \cdot D) + (C \times A) (B \cdot D) \tag{2}$$

which can be seen as

$$ [(A \times B) \cdot C] D = {\bf{M}}D \tag{3}$$

So it seems that proving $(1)$ is equivalent to proving $(2)$. I can prove $(2)$ directly which is given in this post. But I cannot prove $(1)$ directly.

My work

Some computations using Summation Convention and the properties of Kronecker's Delta and Permutation Symbol lead to

$$\eqalign{ & \,\,\,\,\,{\left[ {(A \times B) \otimes C + (B \times C) \otimes A + (C \times A) \otimes B} \right]_{km}} \cr & = {A_i}{B_j}{C_m}{\varepsilon _{ijk}} + {B_i}{C_j}{A_m}{\varepsilon _{ijk}} + {C_i}{A_j}{B_m}{\varepsilon _{ijk}} \cr & = {A_i}{B_j}{C_n}{\delta _{nm}}{\varepsilon _{ijk}} + {B_i}{C_j}{A_n}{\delta _{nm}}{\varepsilon _{ijk}} + {C_i}{A_j}{B_n}{\delta _{nm}}{\varepsilon _{ijk}} \cr & = {A_i}{B_j}{C_n}{\delta _{nm}}{\varepsilon _{ijk}} + {B_j}{C_n}{A_i}{\delta _{im}}{\varepsilon _{jnk}} + {C_n}{A_i}{B_j}{\delta _{jm}}{\varepsilon _{nik}} \cr & = \left( {{\delta _{nm}}{\varepsilon _{ijk}} + {\delta _{im}}{\varepsilon _{jnk}} + {\delta _{jm}}{\varepsilon _{nik}}} \right){A_i}{B_j}{C_n} \cr & = \left( {{\delta _{nm}}{\varepsilon _{ijk}} - {\delta _{im}}{\varepsilon _{njk}} - {\delta _{jm}}{\varepsilon _{ink}}} \right){A_i}{B_j}{C_n} \cr & \mathop = \limits^{???} \left( {{\varepsilon _{ijn}}{\delta _{km}}} \right){A_i}{B_j}{C_n} \cr & = \left( {{A_i}{B_j}{C_n}{\varepsilon _{ijn}}} \right){\delta _{km}} \cr & = \left[ {\left( {A \times B} \right) \cdot C} \right]{\delta _{km}} \cr & = {\left\{ {\left[ {\left( {A \times B} \right) \cdot C} \right]{\bf{I}}} \right\}_{km}} \cr} $$

And the only thing remains to prove is

$${\delta _{nm}}{\varepsilon _{ijk}} - {\delta _{im}}{\varepsilon _{njk}} - {\delta _{jm}}{\varepsilon _{ink}} = {\varepsilon _{ijn}}{\delta _{km}} \tag{4}$$

Is there an elegant way to prove $(4)$ or $(1)$ directly?

  • 1
    $\begingroup$ By the way, the relation $(4)$ is known as a Syzygy (a non-trivial relation between invariant tensors). $\endgroup$ – Oscar Cunningham Nov 26 '15 at 10:13
  • $\begingroup$ @OscarCunningham: Ooops... Thanks for the technology! :) I didn't know that! :) $\endgroup$ – H. R. Nov 26 '15 at 10:16
  • $\begingroup$ @OscarCunningham: Do you know a nice way of proving Syzygies? :) $\endgroup$ – H. R. Nov 26 '15 at 10:19
  • 1
    $\begingroup$ No, sorry. They are studied in the area known as "invariant theory", but all the texts in that area are written in terms of polynomials and algerbraic geometry and so are nigh unreadable to me. I'm sure there is an answer to your question in there if you want to go looking, but don't expect it to be easy. $\endgroup$ – Oscar Cunningham Nov 26 '15 at 10:23

The following argument just occured to me. (I probably half-remembered it from Cvitanovic's Birdtrack book.)

We use the relation


to expand


in two different ways.


$$\begin{align*} \varepsilon_{ani}\varepsilon_{abm}\varepsilon_{bjk}&=(\delta_{nb}\delta_{im}-\delta_{nm}\delta_{ib})\varepsilon_{bjk}\\ &=\delta_{im}\varepsilon_{njk}-\delta_{nm}\varepsilon_{ijk}. \end{align*}$$


$$\begin{align*} \varepsilon_{ani}\varepsilon_{abm}\varepsilon_{bjk}&=\varepsilon_{ani}(\delta_{mj}\delta_{ak}-\delta_{mk}\delta_{aj})\\ &=\delta_{mj}\varepsilon_{kni}-\delta_{mk}\varepsilon_{jni}. \end{align*}$$

Equating these gives


and rearranging gives $(4)$.

  • $\begingroup$ Geeee! Abracadabra! Are you a magician or something? :D That's nice! How it occurred to you!? You have some gifted mind. :) $\endgroup$ – H. R. Nov 26 '15 at 19:15
  • $\begingroup$ Well like I said, I think that it or something like it is in the book by Cvitanovic. I suppose that since the relation $\varepsilon_{iab}\varepsilon_{ipq}=(\delta_{ap}\delta_{bq}-\delta_{aq}\delta_{bp})$ is really the only tool that we have, and the identity we want to prove has only one $\varepsilon$ we can see that any proof has to go via some expression with $3$ $\varepsilon$s. Once you realise that it's not too hard to find the answer. $\endgroup$ – Oscar Cunningham Nov 26 '15 at 19:22
  • $\begingroup$ Is this a joke? :) I mean the name of the book! BirdTracks, doesn't seem to be a mathematics book! I thought you used some humor in your words in your answer! :D $\endgroup$ – H. R. Nov 26 '15 at 19:26
  • 1
    $\begingroup$ See for yourself: http://birdtracks.eu/. It's full name is "Group Theory: Birdtracks, Lie’s, and Exceptional Groups". The word "birdtracks" comes from the diagrams it uses, which look like the patterns a bird might leave in the sand. $\endgroup$ – Oscar Cunningham Nov 26 '15 at 19:29
  • 1
    $\begingroup$ (+1) Thanks for the contribution by the way. This was something I liked. I added a link to your answer. :) $\endgroup$ – H. R. Nov 26 '15 at 19:33

Linear Independent Case

A nice way of proving this is to use non-orthogonal basis for $\mathbb{R}^3$. Hence, consider the following definitions for the non-orthogonal basis

$$\matrix{ {{{\bf{g}}_1} = A} & {{{\bf{g}}_2} = B} & {{{\bf{g}}_3} = C} \cr } \tag{1}$$

and then the dual basis will be

$$\matrix{ {V = \left( {{{\bf{g}}_1} \times {{\bf{g}}_2}} \right) \cdot {{\bf{g}}_3}} \hfill & {} \hfill & {} \hfill \cr {{{\bf{g}}^1} = {{{{\bf{g}}_2} \times {{\bf{g}}_3}} \over V},} \hfill & {{{\bf{g}}^2} = {{{{\bf{g}}_3} \times {{\bf{g}}_1}} \over V},} \hfill & {{{\bf{g}}^3} = {{{{\bf{g}}_1} \times {{\bf{g}}_2}} \over V}} \hfill \cr {{{\bf{g}}^k} = {1 \over {2V}}{\varepsilon ^{ijk}}{{\bf{g}}_i} \times {{\bf{g}}_j}} \hfill & {} \hfill & {} \hfill \cr {{{\bf{g}}_i} \times {{\bf{g}}_j} = V{\varepsilon _{ijk}}{{\bf{g}}^k}} \hfill & {} \hfill & {} \hfill \cr } \tag{2}$$

next, we write the second order identity tensor in the basis ${{\bf{g}}^i} \otimes {{\bf{g}}_j}$ as follows

$${\bf{I}} = I_i^j{{\bf{g}}^i} \otimes {{\bf{g}}_j} \tag{3}$$

then we dot product by ${{\bf{g}}_m}$ from left and by ${{\bf{g}}^n}$ from right and use the property ${{\bf{g}}^i} \cdot {{\bf{g}}_j} = {{\bf{g}}_i} \cdot {{\bf{g}}^j} = \delta _j^i$ to obtain

$${{\bf{g}}_m}{\bf{I}}{{\bf{g}}^n} = I_m^n = \delta _m^n \tag{4}$$

Now, we combine $(2)$, $(3)$, and $(4)$ to get

$${\bf{I}} = I_i^j{{\bf{g}}^i} \otimes {{\bf{g}}_j} = \delta _i^j{{\bf{g}}^i} \otimes {{\bf{g}}_j} = {{\bf{g}}^j} \otimes {{\bf{g}}_j} = {{\bf{g}}^k} \otimes {{\bf{g}}_k}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = {1 \over V}\left( {{1 \over 2}{\varepsilon ^{ijk}}{{\bf{g}}_i} \times {{\bf{g}}_j}} \right) \otimes {{\bf{g}}_k}{\rm{ }} \tag{6}$$

or equivalently

$$\left[ {\left( {{{\bf{g}}_1} \times {{\bf{g}}_2}} \right) \cdot {{\bf{g}}_3}} \right]{\bf{I}} = \left( {{{\bf{g}}_1} \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_3} + \left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes {{\bf{g}}_1} + \left( {{{\bf{g}}_3} \times {{\bf{g}}_1}} \right) \otimes {{\bf{g}}_2} \tag{7}$$

which is the relation $(1)$ in the question with a slightly different notation.

Linear Dependent Case

Consider the case when the vectors $\bf{g}_1$, $\bf{g}_2$, and $\bf{g}_3$ are linearly dependent and hence they cannot form a basis for $\mathbb{R}^3$. This means that there exists real numbers $a$ and $b$ not both zero such that ${{\bf{g}}_1} = a{{\bf{g}}_2} + b{{\bf{g}}_3}$ will hold. This will lead to $V=0$ and turns equation $(7)$ into a trivial identity

$$\eqalign{ & V{\bf{I}} = \left( {{{\bf{g}}_1} \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_3} + \left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes {{\bf{g}}_1} + \left( {{{\bf{g}}_3} \times {{\bf{g}}_1}} \right) \otimes {{\bf{g}}_2} \cr & {\bf{0}} = \left( {\left( {a{{\bf{g}}_2} + b{{\bf{g}}_3}} \right) \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_3} + \left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes \left( {a{{\bf{g}}_2} + b{{\bf{g}}_3}} \right) + \left( {{{\bf{g}}_3} \times \left( {a{{\bf{g}}_2} + b{{\bf{g}}_3}} \right)} \right) \otimes {{\bf{g}}_2} \cr & {\bf{0}} = b\left( {{{\bf{g}}_3} \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_3} + a\left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes {{\bf{g}}_2} + b\left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes {{\bf{g}}_3} + a\left( {{{\bf{g}}_3} \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_2} \cr & {\bf{0}} = - b\left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes {{\bf{g}}_3} - a\left( {{{\bf{g}}_3} \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_2} + b\left( {{{\bf{g}}_2} \times {{\bf{g}}_3}} \right) \otimes {{\bf{g}}_3} + a\left( {{{\bf{g}}_3} \times {{\bf{g}}_2}} \right) \otimes {{\bf{g}}_2} \cr & {\bf{0}} = {\bf{0}} \cr} \tag{8}$$


Relation $(7)$ will hold for every three vectors $\bf{g}_1$, $\bf{g}_2$, and $\bf{g}_3$ in $\mathbb{R}^3$. Finally, replacing ${{\bf{g}}_1}$ by ${{\bf{g}}_i}$, ${{\bf{g}}_2}$ by ${{\bf{g}}_j}$, and ${{\bf{g}}_3}$ by ${{\bf{g}}_k}$ in $(7)$ and doing some further computations will prove identity $(4)$ in the question.

  • $\begingroup$ It seems that you should also check the cases where $A$, $B$, and $C$ are linearly dependant. But then you are doing essentially just as many cases as Jon Mark Perry's answer. $\endgroup$ – Oscar Cunningham Nov 26 '15 at 10:28
  • $\begingroup$ @OscarCunningham: I think this one is more elegant as it is easy to track and also connects some ideas! :) If we have linear dependency we can just check the equality $(1)$ in the question holds by noting the cases of linearly dependency and that $V=0$. I have done something similar in this post $\endgroup$ – H. R. Nov 26 '15 at 10:32
  • $\begingroup$ @OscarCunningham: Added the linear dependent part for your sake to see it is much easier to deal with! :) $\endgroup$ – H. R. Nov 26 '15 at 10:48

This answer uses representation theory. Let $V=\mathbb{R}^3$ be the standard representation of $\mathrm{SO}(3)$, and suppose that $T:V\to V$ is a $\mathrm{SO}(3)$-equivariant map. Schur's lemma would say that $T$ must be a scalar multiple of the identity because $V$ is irreducible, but this is a real representation rather than a complex one so we can't say that automatically.

Instead, observe $T$'s characteristic polynomial is cubic, so it has a real zero, so for some $\lambda$ the operator $T-\lambda\cdot\mathrm{Id}$ has nontrivial kernel containing some line $L$ on which $T$ acts by the scalar $\lambda$. Pick rotations $R,S\in\mathrm{SO}(3)$ so that $V=L\oplus RL\oplus SL$, then $T(R\ell)=RT\ell=\lambda(R\ell)$ for $\ell\in L$ tells us that $T$ acts by the scalar $\lambda$ on $RL$, and similarly for the last line $SL$, so we conclude $T=\lambda\cdot\mathrm{Id}$.

Using the dot product we can identify $V$ with its dual space $V^*$. If $v\in V$, the corresponding dual vector $v^*$ is defined by $v^*(w):=v\cdot w$. In general $V\otimes V^*$ may be identified with $\mathrm{End}(V)$, as representations of $\mathrm{SO}(3)$.

Denote $T_{abc}=(a\times b)\otimes c^*+(b\times c)\otimes a^*+(c\times a)\otimes b^*\in\mathrm{End}(V)$. This defines an antisymmetric trilinear map $V\times V\times V\to\mathrm{End}(V)$, which induces a linear map $\bigwedge^3V\to\mathrm{End}(V)$. In fact notice this map is $\mathrm{SO}(3)$-equivariant. Since $\dim\bigwedge^3V=1$, the map $\mathrm{SO}(3)\to\mathrm{GL}(\bigwedge^3 V)\cong\mathbb{R}^\times$ must be trivial because $\mathrm{SO}(3)$ is compact and connected, so $\bigwedge^3V$ is the trivial representation and its image in $\mathrm{End}(V)$ must be contained in the space of $\mathrm{SO}(3)$-equivariant endomorphisms, so $T=\lambda\cdot\mathrm{Id}$ for some $\lambda$.

Using the fact $\mathrm{tr}(v\otimes w^*)=v\cdot w$, we see

$$3\lambda=\mathrm{tr}(T_{abc})=(a\times b)\cdot c+(b\times c)\cdot a+(c\times a)\cdot b=3\,(a\times b)\cdot c$$

and therefore $T_{abc}=((a\times b)\cdot c)\,\mathrm{Id}$.

  • $\begingroup$ (+1), As the technology was out of my league I couldn't understand much! :) However, I am enjoying to see a nice different and compact solution! :) $\endgroup$ – H. R. Nov 27 '15 at 7:31

Using clifford algebra, the identity (4) can be rewritten as (using vectors $m, n, i, j, k$)

$$(m \cdot i)(j \wedge k \wedge n) + (m \cdot j)(i \wedge k \wedge n) + (m \cdot k)(i \wedge j \wedge n) - (m \cdot n)(i \wedge j \wedge k) = 0$$

$m$ is arbitrary, and therefore it can be canceled to yield

$$i(j \wedge k \wedge n) + j (i \wedge n \wedge k) - k(j \wedge i \wedge n) - n(i \wedge j \wedge k) = 0$$

As a geometric identity, this reads, "Given a set of four vectors, form a volume from each combination of three. Then, find the plane orthogonal to the fourth and weight it such that that plane and that vector span the volume with the magnitude of that volume. The four such oriented, weighted planes formed from each combination add to zero."

We can solve this problem (prove the identity) by using clifford products and grade projection. Associativity of the product operation allows us to quickly derive true relations from grade projection, and these relations involve the terms that appear in the desired identity.

We'll do this:

$$\begin{align*} \langle ijkn \rangle_2 &= \color{red}{i(j \wedge k \wedge n)} + i \wedge [j (k \cdot n) - k (j \cdot n) + n(j \cdot k)] \\ &= \color{red}{(i \wedge j \wedge k)n} + [i(j \cdot k) + k (i \cdot j) - j(i \cdot k)] \wedge n\\ \langle jink \rangle_2 &= \color{red}{j (i \wedge n \wedge k)} + j \wedge [i(n \cdot k) + k (i \cdot n) - n(i \cdot k)] \\ &= \color{red}{(j \wedge i \wedge n) k} + [j(i \cdot n) + n(j \cdot i) - i(j \cdot n)] \wedge k \end{align*}$$

The requisite terms for identity (4) are in red (mind that with the wedge products, there are apparent, but not substantial, differences in minus signs).

Rearrange these identities to yield equations of the form

$$\begin{align*} i(j \wedge k \wedge n) - (i \wedge j \wedge k)n + \ldots &= 0 \\ j(i \wedge n \wedge k) - (j \wedge i \wedge n)k + \ldots &= 0 \end{align*}$$

Then add those two equations to arrive at the desired identity. All the other terms (...) will cancel: the $i \wedge n$ terms and $j \wedge k$ terms cancel during the subtraction; the other terms cancel during the addition.

  • $\begingroup$ Thanks but I cannot really understand! :) What is Clifford algebra? What is $\wedge$? Is it the cross product? :) And why $i.m=\delta_{im}$ holds if $i$ and $m$ are two arbitrary vectors? $\endgroup$ – H. R. Nov 30 '15 at 16:30
  • 1
    $\begingroup$ A clifford algebra uses a vector multiply operation--the "clifford product" that I have used here. This operation obeys $mm = |m|^2$ and $ijk = (ij)k = i(jk)$. The resulting algebra has as its elements multivectors, some of which correspond to planes or volumes instead of lines (as vectors do). - The symbol $\wedge$ here should be read as "wedge". The wedge product of two vectors is a 2-vector or bivector and corresponds to a plane. The wedge product of three vectors corresponds to a volume. These can be written in terms of the clifford product. - Note that $i^a m^b \delta_{ab} = i \cdot m$. $\endgroup$ – Muphrid Nov 30 '15 at 16:40
  • $\begingroup$ (+1): Thanks! :) I think that I have some lack of knowledge about this; however, it is nice to see many different solutions for this problem! :) $\endgroup$ – H. R. Nov 30 '15 at 16:44


$${\delta _{nm}}{\varepsilon _{ijk}} - {\delta _{im}}{\varepsilon _{njk}} - {\delta _{jm}}{\varepsilon _{ink}} = {\varepsilon _{ijn}}{\delta _{km}}$$


$${\delta _{nm}}{\varepsilon _{ijk}} = {\delta _{im}}{\varepsilon _{njk}} + {\delta _{jm}}{\varepsilon _{ink}} + {\delta _{km}}{\varepsilon _{ijn}}$$

makes it easier to see the symmetry. Assuming $n,m,i,j,k\in\{1,2,3\}$ we can easily prove the case for $i=j=k$, namely that every $\varepsilon=0$.

Assume two of $i,j,k$ are equal, say $i=j$. Then the left-hand side is $0$, and the $3^{rd}$ term on the right-hand side is zero. But we also have that:

$${\delta _{im}}{\varepsilon _{njk}} + {\delta _{jm}}{\varepsilon _{ink}}$$

$$= {\delta _{im}}{\varepsilon _{njk}} + {\delta _{im}}{\varepsilon _{jnk}}$$

$$= {\delta _{im}}{\varepsilon _{njk}} - {\delta _{im}}{\varepsilon _{njk}}$$


And so the equations holds for this case as well.

So we need to consider $i,j,k$ distinct.

If $n\ne m$, we have the LHS disappearing, and only one term to consider on the RHS, for example if $i=m$ so that $\delta_{im}=1$. However as $n\ne m$ and the $i,j,k$ are distinct, we have that either $j=n$ or $k=n$ and so this term is $0$ too.

Finally, if $n=m$, the LHS is $\varepsilon_{ijk}$ and if again for example $i=m$ then we have:

$${\delta _{im}}{\varepsilon _{njk}}$$

$$=\varepsilon _{njk}$$

But $n=m=i$ so:

$$=\varepsilon _{ijk}$$

  • $\begingroup$ (+1) Thanks for the contribution. :) You have proved the identity $(4)$ by investigating approximately all the cases! There are lots of cases, namely $3^5=243$. :) It is some kind of brute force and I really don't like proving this way! Thanks again by the way. :) $\endgroup$ – H. R. Nov 25 '15 at 22:19

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