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I am working on a problem relating to Lyapunov exponents of products of random matrices, and this has led me to the following question which I suspect is best approached using techniques outside my area of expertise:

Let $\tau \colon SL_2^\pm(\mathbb{R})^k \to \mathbb{R}$ be a function which has the form $$\tau(A_1,\ldots,A_k):=\mathrm{tr }\left(A_{i_n}A_{i_{n-1}}\cdots A_{i_2}A_{i_1}\right)$$ for some fixed indices $i_1,i_2,\ldots,i_n \in \{1,\ldots,k\}$. Can $\tau$ have a local maximum or a local minimum at one of its zeros?

My main area of mathematical expertise is analysis, so I apologise if this is actually very easy to solve using tools with which I am unfamiliar. Since I am not sure which aspects of the problem are most significant in its solution I have taken a somewhat scattershot approach to tagging: please feel free to adjust the tags if you feel that they are inappropriate.

Some remarks on the question:

  • Local maxima and minima are of course not strict, since if $B \in SL_2(\mathbb{R})$ is close to the identity then $(B^{-1}A_1B,\ldots,B^{-1}A_kB)$ is close to, and typically distinct from, $(A_1,\ldots,A_k)$ but is taken to the same value by $\tau$.
  • It is certainly possible for a function of this type to have a local minimum. For example, if $k=1$, $\tau(A_1):=\mathrm{tr }A_1^2$ then a local minimum is achieved when $A_1$ has trace zero and determinant $-1$. The value of $\tau$ at the minimum in this case is $2$.
  • A solution which treated only the case $k=2$, but in which $n$ and $i_1,\ldots,i_n \in \{1,2\}$ were allowed to be arbitrary, would still be interesting to me.
  • I would find a complete treatment of the case $\det A_{i_n} \cdots A_{i_1}=1$ mildly interesting, but my main interest is in the case where the determinant of the product is negative.

Updated to add: Will Sawin shows below that if $(A_1,\ldots,A_n)$ is a critical point of $\tau$ with respect to differentiation in the variable $A_r$, and is also a zero of $\tau$, then $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} A_{i_\ell} \cdots A_{i_1}A_{i_n}\cdots A_{i_{\ell+1}}=0$$ and this equation is impossible if $(A_1,\ldots,A_k) \in SL_2(\mathbb{R})^k$. Using a slightly different argument suggested by Will's answer I can show that a tuple $(A_1,\ldots,A_k) \in SL_2^\pm(\mathbb{R})^k$ which satisfies $\tau(A_1,\ldots,A_k)=0$ is a critical point of $\tau$ if and only if for each $r \in \{1,\ldots,k\}$ $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left((A_{i_n}\cdots A_{i_{\ell+1}})X(A_{i_n}\cdots A_{i_{\ell+1}})^{-1}+(A_{i_\ell}\cdots A_{i_{1}})^{-1}X(A_{i_\ell}\cdots A_{i_{1}}))\right)=0$$ for every traceless matrix $X$. Details are given below as an answer. This result implies in particular that if any symbol $r \in \{1,\ldots,k\}$ occurs exactly once in the word $i_1\cdots i_n$, then $\tau$ does not have a zero which is also a critical point, a result which is clearly also implied by Will's work.

It is at this stage an open question whether $\tau$ can have critical points which are also zeros.

Here is an example which shows that a zero of $\tau$ can be a critical point with respect to directional derivatives in a single co-ordinate. Define $$B_1:=\left(\begin{array}{cc}1&0\\0&1\end{array}\right),\qquad B_2:=\left(\begin{array}{cc}0&1\\1&0\end{array}\right),\qquad B_3:=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right).$$ It can be checked by hand that $B_3e^{tX}B_1B_2e^{tX}B_1$ is traceless for all $t \in \mathbb{R}$ when $X$ is any of the traceless matrices $$\left(\begin{array}{cc}1&0\\0&-1\end{array}\right),\qquad \left(\begin{array}{cc}0&1\\0&0\end{array}\right),\qquad \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$ and in particular in this case the derivative of $\tau$ in the first variable at $(B_1,B_2,B_3)$ is zero. One may easily check by hand that $B_3B_1B_2B_1$ is traceless and hence is a zero of $\tau(A_1,A_2,A_3):=A_3A_1A_2A_1$. This triple is not a critical point with respect to directional derivatives in the other variables.

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    $\begingroup$ So the answer "should" be no, right? $\endgroup$ – Anthony Quas Jun 21 '14 at 1:32
  • $\begingroup$ @AnthonyQuas: Yes, the desired answer would be "no". $\endgroup$ – Ian Morris Jun 21 '14 at 9:57
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No. In fact there are no critical points of trace $0$. We will establish a contradiction by assuming the trace and derivative are $0$. The tangent space of $SL_2(R)$ at $A$ is isomorphic to the Lie algebra $sl_2(R)$, with the isomorphism defined by the tangent vector $X$ pointing in the direction of the matrix $(1+X)A$.

We are going to take the derivative in direction $X$ of the $i$th copy of $SL_2$. By the product rule, this is the sum over all $m$ such that $i_m=i$ of $tr(A_{i_n} \dots A_{i_{m+1}} X A_{i_m} \dots A_{i_1})$. By the rotation property of the trace, this is the same as the sum over all such $m$ of $tr(X A_{i_m} \dots A_{i_1}A_{i_n} \dots A_{i_{m+1}})$.

$A_{i_m} \dots A_{i_1}A_{i_n} \dots A_{i_{m+1}}$ is conjugate to $A_{i_n}\dots A_{i_1}$, so it is a matrix of trace $0$. So the sum of the $A_{i_m} \dots A_{i_1}A_{i_n} \dots A_{i_{m+1}}$ is a matrix of trace $0$, hence an element of $sl_2(\mathbb R)$. Because we are at a local maximum, its product under the Killing form with every element $X \in sl_2(\mathbb R)$ is $0$. Because the Killing form is nondegenerate, it is $0$.

So we have some matrix $A_n\dots A_1$ in $SL_2(\mathbb R)$ of trace $0$ such that summing up a bunch of conjugates of it produces the $0$ matrix. I claim this is impossible! To see why, note that the set of such matrices, defined by the equations $a+d=0$, $ad-bc=1$, is a two-sheeted hyperboloid with two connected components. Since $SL_2(\mathbb R)$ is connected, its action by conjugation cannot switch the two components (or use the fact that the components correspond to orientations). Visually, it is clear that summing up members of either component never produces the $0$ matrix. (In one component, $b-c \geq 2$ and in the other $b-c \leq -2$) So we have a contradiction.

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  • $\begingroup$ This is a good answer which has helped me a lot to think about the problem. However, I defined $\tau$ on $SL_2^\pm(\mathbb{R})^k$ and this answer treats only $SL_2(\mathbb{R})^k$. Consider the pair in $SL_2^\pm(\mathbb{R})^2$ given by $$A_1:=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right),\qquad A_2:=\left(\begin{array}{cc}0&1\\1&0\end{array}\right).$$ We have $A_1,A_2,A_1A_2 \in sl_2(\mathbb{R})$, $A_1A_2 \neq 0$, and $A_1A_2+A_2A_1=0$, which as you note above is impossible when $A_1,A_2 \in SL_2(\mathbb{R})$. $\endgroup$ – Ian Morris Jun 21 '14 at 18:22
  • $\begingroup$ Note that the pair $A_1,A_2$ defined in my comment above is not a critical point of $\tau(B_1,B_2):=B_1B_2$, since if $R_\theta$ denotes the matrix of anticlockwise rotation through angle theta, then $A_1 R_\theta A_2$ has trace $2\sin\theta$ and so the derivative of $\tau$ is nonzero. So I think that your answer is on the right track, but that a further argument is required to show that the derivative is nonzero. $\endgroup$ – Ian Morris Jun 21 '14 at 18:31
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Here are some thoughts of my own which derive from Will Sawin's answer (which solved the problem in the case $(A_1,\ldots,A_k) \in SL_2(\mathbb{R})^k$). Let us suppose that $(A_1,\ldots,A_k) \in SL_2^\pm(\mathbb{R})$ is a critical point of $\tau$ such that $\tau(A_1,\ldots,A_k)=0$. I claim that for every $r \in \{1,\ldots,k\}$, the equation $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left((A_{i_n}\cdots A_{i_{\ell+1}})X(A_{i_n}\cdots A_{i_{\ell+1}})^{-1}+(A_{i_\ell}\cdots A_{i_{1}})^{-1}X(A_{i_\ell}\cdots A_{i_{1}}))\right)=0$$ must be satisfied for all traceless matrices $2 \times 2$ real matrices $X$. Furthermore this is a necessary and sufficient condition for a zero $(A_1,\ldots,A_k)$ of $\tau$ to be a critical point of $\tau$.

To see this let us fix $r$ and $X$ and define $\sigma:=\det A_{i_n}\cdots A_{i_1} \in \{\pm1\}$. We suppose that $(A_1,\ldots,A_k)$ is both a zero and a critical point of $\tau$. If $(B_1,\ldots,B_k)$ is in the same connected component of $SL_2^\pm(\mathbb{R})^k$ as $(A_1,\ldots,A_k)$ then by the Cayley-Hamilton theorem $$(B_{i_n}\cdots B_{i_1})^2 - (\tau(B_1,\ldots,B_k))B_{i_n}\cdots B_{i_1} +\sigma I = 0.$$ Let us differentiate this equation along the one-parameter family $t \mapsto (A_1,\ldots,e^{tX}A_r,\ldots,A_k)$ at the value $t=0$. Since by hypothesis $\tau$ and its derivative are zero at $(A_1,\ldots,A_k)$, the only nonzero terms in the derivative of the Cayley-Hamilton identity are those arising from the first term, namely $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} \left(A_{i_n}\cdots A_{i_{\ell+1}}XA_{i_\ell} \cdots A_{i_1}\right)A_{i_n}\cdots A_{i_1}$$ and $$\sum_{\substack{1 \leq \ell \leq n\\ i_\ell=r}} A_{i_n}\cdots A_{i_1}\left(A_{i_n}\cdots A_{i_{\ell+1}}XA_{i_\ell} \cdots A_{i_1}\right)$$ and the total of these two sums must equal zero. Again by the Cayley-Hamilton theorem we have $$(A_{i_n}\cdots A_{i_1})^2=-\sigma I$$ so in particular $$A_{i_n}\cdots A_{i_1}A_{i_n}\cdots A_{i_{\ell+1}}=-\sigma (A_{i_\ell}\cdots A_{i_1})^{-1}$$ and $$A_{i_\ell}\cdots A_{i_1}A_{i_n}\cdots A_{i_1}=-\sigma (A_{i_n}\cdots A_{i_{\ell+1}})^{-1}.$$ Substituting these identities into the results obtained thus far we obtain the claimed identity. Note that all the steps of this argument are reversible.

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