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I recently came across this curious fact in some calculations with the strain tensor in fluid mechanics:

Let $A$ be an antisymmetric 3 by 3 matrix and $S$ be a traceless symmetric 3 by 3 matrix. Then it is easy to see that $AS+SA$ is an antisymmetric 3 by 3 matrix. $A$ corresponds to a 3-dimensional vector $v$ in the following way: for any 3-dimensional vector $w$ we have $Aw=v\times w$. When the same procedure is applied to $AS+SA$, it will produce the vector $-Sv$.

I have two ways to prove this. The first is to write everything in terms of indices, and watch the big chunk mysteriously fizzles away after massive cancellation. The other way is representation-theoretic: it boils down to the fact that, in terms of $SO(3)$-representations, $3\otimes 5=3\oplus5\oplus7$, and everything except the constant factor follows from Schur's Lemma.

The representation-theoretic proof also says that the projection $3\otimes 5\to 3$ is "natural" in some sense. But now I have two descriptions of this projection, so is there a natural way (without indices, but also more concrete than abstract representation-theoretic nonsense) to see that they agree up to the minus sign?

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    $\begingroup$ Hm, $(AS+SA)^t=S^tA^t+A^tS^t=-SA-AS$, where I am wrong? $\endgroup$
    – Fedor Petrov
    Commented Oct 11, 2017 at 7:02
  • $\begingroup$ What if $A$ is the zero matrix? Then how could $AS+SA=0$ produce the vector $-Sw$? $\endgroup$
    – Somos
    Commented Oct 11, 2017 at 11:15
  • $\begingroup$ (I have corrected the typo: should read $-Sv$ instead of $-Sw$) $\endgroup$
    – Qfwfq
    Commented Oct 11, 2017 at 12:34
  • $\begingroup$ @FedorPetrov You're absolutely right in showing that $AS+SA$ is antisymmetric, so it can be used to produce a vector in the same way as $A$. $\endgroup$
    – Fan Zheng
    Commented Oct 11, 2017 at 14:24
  • $\begingroup$ ah, the statement you ask about is in the next paragraph! This confused me. $\endgroup$
    – Fedor Petrov
    Commented Oct 11, 2017 at 14:35

1 Answer 1

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Your claim can be restated and generalised slightly as $$S(v\times w)+(Sv)\times w+v\times(Sw)=\text{trace}(S)v\times w$$ for all symmetric $S$ and all $v,w\in\mathbb{R}^3$. Just as with the trace-free case, it is straightforward to check the identity by a large and unilluminating calculation.

Another approach is as follows. We have an orthonormal basis of eigenvectors $e_i$ with eigenvalues $s_i$ say and the cross product of any two of them will be the third one up to sign. It is then easy to check the identity when $v=e_i$ and $w=e_j$ (noting that both sides are zero if $i=j$). The general case then follows by bilinearity.

My guess is that there is a better, more direct proof but I cannot see one at the moment. I agree that this is a curious fact, and I am surprised that I have never seen it before.

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  • $\begingroup$ Thanks for the nice explanation. Also, note both sides are antisymmetric w.r.t exchange of $v$ and $w$, so only half of the cases $i\neq j$ needs to be checked, and by cyclic permutation of the coordinates, there is essentially a single case left, which boils down to $s_1+s_2+s_3=\text{trace}(S)$. $\endgroup$
    – Fan Zheng
    Commented Oct 11, 2017 at 14:28
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    $\begingroup$ Taking inner products with a test vector $u$ and using the symmetry of $S$, your identity follows from the identity $Su \wedge v \wedge w + u \wedge Sv \wedge w + u \wedge v \wedge Sw = \mathrm{trace}(S) (u \wedge v \wedge w)$, which is the infinitesimal form of $Tu \wedge Tv \wedge Tw = \mathrm{det}(T) (u \wedge v \wedge w)$, which is the usual volume-theoretic interpretation of the determinant. $\endgroup$
    – Terry Tao
    Commented Oct 11, 2017 at 16:10
  • $\begingroup$ @TerryTao Very nice insight! And it's amusing to see the word "volume-theoretic" (a stripped down version of "measure-theoretic"?) $\endgroup$
    – Fan Zheng
    Commented Oct 11, 2017 at 16:22
  • $\begingroup$ Lebesgue measure is unsigned. Volume forms are signed. Measure theory can interpret $|\mathrm{det}(T)|$ but not $\mathrm{det}(T)$. $\endgroup$
    – Terry Tao
    Commented Oct 11, 2017 at 16:28
  • $\begingroup$ @TerryTao Thanks for clarification. $\endgroup$
    – Fan Zheng
    Commented Oct 17, 2017 at 0:50

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