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Jun 21, 2014 at 18:31 comment added Ian Morris 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.
Jun 21, 2014 at 18:22 comment added Ian Morris 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 $A_1,A_2 \in SL_2(\mathbb{R})$.
Jun 21, 2014 at 12:29 history answered Will Sawin CC BY-SA 3.0