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Consider the space of $n \times n$ complex matrices with metric equal to the distance in the Frobenius norm so that $d(A,B) \equiv \| A - B \| \equiv \sqrt{Tr[(A-B)^{\dagger}(A-B)]}$. I want to know the volume of a ball defined as $B_{\epsilon}(U_0) \equiv \{U: \|U - U_0\| \leq \epsilon\}$. What are some appropriate notions of volume and what would each mean? (Answers that don’t assume previous knowledge of measure theory would be helpful).

What if I want to compute this based on some other norm distance? What is the general way to think about it?

Also, what if I want to restrict $U$ to unitary matrices, so that for example I have a ball $S_{\epsilon}(U_0) \equiv \{U \in U(n): \|U - U_0\| \leq \epsilon\}$? What if I further restrict it to $V_{\epsilon}(U_0) \equiv \{U \in SU(n): \|U - U_0\| \leq \epsilon\}$?

Note: please help with tags ...

Sherif F.
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    This might have to do with Haar measure, but I am not certain. – angryavian Dec 23 '17 at 04:41
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    What does "volume" mean in this context? If you mean Lebesgue measure on $\mathbb C^{n^2}$ the answer to the main question is $\epsilon^{2n^2}$ times an $n$-dependent expression involving the $\Gamma$ function. – kimchi lover Dec 23 '17 at 14:09
  • @kimchilover so I’m not familiar with measure theory. A brief explanation of what volume might mean say for Lebesgue and/or Haar measures (simply being common ones) might help as part of an answer, and/or some choice of whatever might be a meaningful notion of volume for this context. Thanks! – Sherif F. Dec 23 '17 at 16:08
  • @kimchilover I edited the question to add this clarification: what are some appropriate notions of volume and what would each of them basically “mean”? – Sherif F. Dec 23 '17 at 16:11
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    In 1 dimension we have length, in 2 we have area, in 3 we have ordinary volume and in general $n$ dimensional Euclidean space we have a corresponding $n$-dimensional analog. This is Lebesgue measure. You can learn more at https://en.wikipedia.org/wiki/Lebesgue_measure and https://en.wikipedia.org/wiki/Volume_of_an_n-ball . – kimchi lover Dec 23 '17 at 16:18
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    Notice that the frobenius norm is just the euclidean norm on the matrix entries. – Nightgap Dec 25 '17 at 02:41

1 Answers1

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For measure theory on groups the notion of multiplication is far more important than the one of distance.

In what follows I will always assume that translation in your group is left multiplication, otherwise if you are just summing matrices, your space is isometric to $\mathbb{R}^{2n}$ and the answer is trivial.

Ingredients for the proof have all been mentioned in comments. First you can identify your group with $\mathbb{R}^{2n}$ endowed with the suitable multiplication. This is a Lie group, and the topology is the one induced by the Euclidean metric (see comments above). Thus, there exists a left Haar measure, which up to multiplication with a smooth function (the so called modular function) is the Lebesgue measure. Note that despite the fact that your group is connected and simply connected, it is not true that the modular function is constant (if I am not wrong in our case, it should be the square root of the determinant, i.e if $\mu$ is the left Haar measure, then $d\mu(X)=\sqrt{\det(X)}dX$). Thus the measure of your set should really be $\mu(B_\epsilon(U_0))$, which is a just a matter of mere computation. Note that this volume will depend even on the centre since your distance is not left invariant (i.e. invariant under left multiplication).

Sherif F.
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Diesirae92
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  • I awarded the bounty for this since it's the only answer and it addresses the question. I wish it would flesh out the concepts a little bit more and would justify some of the claims and statements that it makes. Thanks! – Sherif F. Jan 01 '18 at 02:01