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Fact: Let $U$ and $V$ be two $ n \times n$ matrices with determinant $ 1.$ Assume that $S_1,S_2,....S_m$ are linearly independent $n \times n$ matrices such that $U^{-1}S_iU$ and $V^{-1}S_iV$ are skew-symmetric for all $i=1,m.$ If $m \geq {n \choose 2}$ then $$ U=OV$$ for $O$ orthogonal matrix.

Can we draw the same conclusion for some values $m < {n \choose 2}?$

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  • $\begingroup$ Related to the question: Assume $\nabla $ is a connection in a vector bundle over the manifold $M$. Assume the dimension of the fiber is $n$ and let $\Omega$ be its curvature matrix in a local frame. Chose $\theta_i,$ $i=1,m$ a local frame of two forms on $M$ and define the matrices $S_i$ as $$ \Omega=\sum_{i=1}^m \theta_i S_i.$$ Then there is an algorithm to determine if the connection is locally metric if $n=2$ or $n>2$ and there are at least ${n-1 \choose 2}+1$ linearly independent matrices among the $S_1,S_2,....S_m$ $\endgroup$
    – Mike Cocos
    Commented Nov 30, 2019 at 22:26

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I will assume, though you didn't say, that the ground field is $\mathbb{R}$. (For all I know, the argument below might fail when the ground field is finite, etc.)

Yes, when $n>2$, it works for $m > {{n-1}\choose2}$, but not for $m = {{n-1}\choose2}$.

The reason is the following: Let $W$ be the span of the matrices $S_1,\ldots,S_m$. Your hypothesis is that, under conjugation by $U^{-1}$, $W$ gets carried into the skew-symmetric matrices, i.e., the Lie algebra of $\mathfrak{so}(n)$, or, equivalently, $W$ is a subspace of the Lie subalgebra $$ L = \{\ UxU^{-1}\ |\ x\in \mathfrak{so}(n)\ \}\subset \mathfrak{sl}(n), $$ which is, of course, isomorphic to $\mathfrak{so}(n)$. Let $K\subset L$ be the subalgebra generated by $W$. Because conjugation by $U$ is an automorphism of the Lie algebra $\mathfrak{sl}(n)$, your hypothesis implies that conjugation by $U^{-1}$ carries $K$ into a subalgebra of $\mathfrak{so}(n)$. Hence, if $K=L$, then your hypothesis implies that conjugation by $U^{-1}$ carries $L$ to $\mathfrak{so}(n)$. Since $L$ has dimension $n\choose 2$, your original argument for $m={n\choose2}$ finishes the job.

Now, when the ground field is $\mathbb{R}$, it's a fact that, except when $n=4$, $\mathfrak{so}(n)$ has no proper subalgebras of dimension greater than ${n-1}\choose2$, so, with the hypothesis $m>{{n-1}\choose2}$, one gets $K=L$, and the above argument finishes the proof. When $n=4$, one has to do a separate argument, because $\mathfrak{so}(4)$ does have proper subalgebras of dimension $4$, for example $\mathfrak{u}(2)$. However, this is the only case, and it is easily verified that, when $n=4$ and $K\simeq\mathfrak{u}(2)$, then you still get the desired result.

Meanwhile if $m={{n-1}\choose2}$, then we can take $W = \mathfrak{so}(n{-}1)$ and find many pairs of matrices $U$ and $V$ with determinant 1 that carry $W$ into $\mathfrak{so}(n)$ but that do not differ by an orthogonal matrix.

Added remark: If one changes the question slightly and asks what is the smallest $m$ for which there exist $n$-by-$n$ matrices $S_1,\ldots, S_m$ that can be simultaneously skew-symmetrized by conjugation and are such that any two unimodular matrices $U$ and $V$ that conjugate the $S_i$ into skew-symmetric matrices must differ by an orthogonal matrix, then the answer (for $n>2$) is $m=2$. This is because, for $n>2$, the Lie algebra $\mathfrak{so}(n)$ can be generated by any two generically chosen elements.

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  • $\begingroup$ Thank you Professor Bryant! Always to the point. So it turns out that if we take a a local frame $\sigma=(\sigma_1, \sigma_2,...,\sigma_m )$ and calculate the curvature matrix $\Omega$ with respect to this frame and then look at the matrices $S_{ij}$(with real smooth functions as entries) defined by the equation $$ \Omega=\sum_{i<j} (\sigma^i \wedge \sigma^j) S_{ij} $$ and if there are ${n-1\choose 2}$ linearly independent among them then the algorithm works. In particular since for $n=2$ there's only one matrix in that family the algorithm should work as well. $\endgroup$
    – Mike Cocos
    Commented Nov 30, 2019 at 22:05
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    $\begingroup$ @MikeCocos: Actually, you need more than ${n-1}\choose2$ linearly independent matrices in order for this argument to work. However, your argument still doesn't address how you are going to get the matrices $U$ and $V$ to be unimodular. In other words, you still have to do an integration to determine the volume form of the metric. It can't be determined algebraically (and it may not be globally possible), as I pointed out in my example. $\endgroup$
    – Robert Bryant
    Commented Dec 1, 2019 at 9:53
  • $\begingroup$ I am not sure what you mean by getting $U$ and $V$ to be unimodular? I am only interested in the local problem. If $U$ and $V$ skew-symmetrize the curvature matrix and since they are non singular just dividing by their determinant will make them unimodular. I am not even pretending to be trying to solve the problem on the whole manifold. All I am interested in is a practical way( using local frames) to determine if a connection is LOCALLY metric. I believe that if there are more than $ {n-1 \choose 2}$ linearly independent matrices among the $S_{ij}$ then algorithm works $\endgroup$
    – Mike Cocos
    Commented Dec 1, 2019 at 23:06
  • $\begingroup$ A possible obstruction for the global metrizability of a connection in any vector bundle is obtained by calculating the Euler class of the connection. The Euler class of any locally metric connection is well defined via the Pfaffian of its curvature matrix. If its Euler class doesn't equal the Euler characteristic of the manifold then the connection cannot be globally metric. I believe that calculating the Euler class of the connection is much simpler than calculating its holonomy group. $\endgroup$
    – Mike Cocos
    Commented Dec 1, 2019 at 23:15
  • $\begingroup$ I hope I'm not becoming annoying but there are more things related to the local metrizability. On an affinely flat manifold one can deform the flat connection $\nabla$ into any globally metric connection $D$ by the standard formula $$\nabla(t)=(1-t)\nabla+tD$$ If one can show that $\nabla(t)$ is locally metric for all $t$ then the Euler class of the manifold equals the Euler class of the flat connection and therefore is zero. This is the main reason I got interested in locally metric connections. $\endgroup$
    – Mike Cocos
    Commented Dec 1, 2019 at 23:24

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