For a problem I'm working on, it would be helpful to be able to calculate $E[M M^T]$, where the expectation value is taken over $SL_N(\mathbb{R})$. Is this well defined? I suspect that if it is, the result is a multiple of $I_N$ (sign-flips should kill off-diagonal terms, I think). Is that in fact the case, and if so what is the constant of proportionality?
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To take an expectation you need a probability measure. $SL_n(\mathbb{R})$ is a noncompact Lie group, so it doesn't admit a left-invariant probability measure. (It admits, up to scale, a unique left-invariant measure called left Haar measure with respect to which $SL_n(\mathbb{R})$ itself has infinite measure, so it cannot be rescaled to a probability measure.) So what probability measure is intended? – Qiaochu Yuan Dec 12 '17 at 06:11
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I know about the Haar measure, and that $SL_n(\mathbb{R})$ has infinite measure under it. That doesn't prevent an expectation value from being defined -- it just means that most expectation values will not be well defined. To be more precise, let ${U_i}$ be a sequence of balls around the origin of increasing radius under the standard metric, such that $\mu(U_i) < \infty$ and $\cup U_i = SL_n(\mathbb{R})$, then define $E_i[f(g)] = \int_{U_i} d\mu(g) f(g) / \mu(U_i)$. Does the sequence $E_i[MM^T]$ converge, and if so, to what? – Craig Dec 12 '17 at 14:37
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And yes, generally this will only work if $f(g) \in L^2(SL_n(\mathbb{R}))$, but one can hope for conditional convergence or some sort of resummation... – Craig Dec 12 '17 at 16:04
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1Try it for $N=2$. – anon Dec 12 '17 at 16:36
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Diverges, right, should have done that to begin with... – Craig Dec 12 '17 at 17:56