For example, in the Gaussian symplectic ensemble (GSE), one can calculate the following expectation values: \begin{align*} \mathbb{E}[ \mathrm{tr} 1 ] &= 2 n\\ \mathbb{E}[ \mathrm{tr} (H^{2}) ] &= n (2 n-1)\\ \mathbb{E}[ \mathrm{tr} (H^{4}) ] &= \frac{1}{2} n \left(8 n^2-10 n+5\right)\\ \mathbb{E}[ \mathrm{tr} (H^{6}) ] &= \frac{1}{4} n \left(40 n^3-88 n^2+104 n-41\right)\\ \mathbb{E}[ \mathrm{tr} (H^{8}) ] &= 28 n^5-93 n^4+187 n^3-\frac{345 n^2}{2}+\frac{509 n}{8}\\ \mathbb{E}[ \mathrm{tr} (H^{10}) ] &=\frac{1}{16} n \left(1344 n^5-6176 n^4+18320 n^3-28600 n^2+24286 n-8229\right)\\ \mathbb{E}[ \mathrm{tr} (H^{12}) ] &=\frac{1}{32} n \left(8448 n^6-50752 n^5+204768 n^4-470080 n^3+668592 n^2-516958 n+166377\right) \end{align*} In a previous blog post I calculated these expectation values in the Gaussian orthogonal ensemble (GOE): \begin{align*} \mathbb{E}[ \mathrm{tr} 1 ] &= 2 n\\ \mathbb{E}[ \mathrm{tr} (H^{2}) ] &= n (2 n+1)\\ \mathbb{E}[ \mathrm{tr} (H^{4}) ] &= \frac{1}{2} n \left(8 n^2+10 n+5\right)\\ \mathbb{E}[ \mathrm{tr} (H^{6}) ] &= \frac{1}{4} n \left(40 n^3+88 n^2+104 n+41\right)\\ \mathbb{E}[ \mathrm{tr} (H^{8}) ] &= 28 n^5+93 n^4+187 n^3+\frac{345 n^2}{2}+\frac{509 n}{8}\\ \mathbb{E}[ \mathrm{tr} (H^{10}) ] &=\frac{1}{16} n \left(1344 n^5+6176 n^4+18320 n^3+28600 n^2+24286 n+8229\right)\\ \mathbb{E}[ \mathrm{tr} (H^{12}) ] &=\frac{1}{32} n \left(8448 n^6+50752 n^5+204768 n^4+470080 n^3+668592 n^2+516958 n+166377\right) \end{align*} If one formally replaces \( n \) by \( -n \) in the expectation values in the GSE, one gets the expectation values in the GOE (up to sign). With a formula one can write \begin{equation} \mathbb{E}[ \mathrm{tr} (H^{2 p}) ]_{\mathrm{GSE}(2n)} =(-1)^{p+1} \mathbb{E}[ \mathrm{tr} (H^{2 p}) ] _{\mathrm{GOE}(-2n)} \end{equation} This is a duality between the Gaussian orthogonal ensemble and the Gaussian symplectic ensemble. I find this a bizarre formula because the natural interpretation of the right hand side seems to involve matrices of "negative dimension". Nevertheless, this connection between the orthogonal group and the symplectic group shows up in various places in mathematics and physics. For example, one has proven that \( Sp(2N) \) and \( SO(−2N) \) gauge theories are equivalent (see reference below).

Similary, I can calculate expectation values in the Gaussian unitary ensemble: \begin{align*} \mathbb{E}[ \mathrm{tr} (H^{2}) ] &= n^2\\ \mathbb{E}[ \mathrm{tr} (H^{4}) ] &= 2 n^3+n\\ \mathbb{E}[ \mathrm{tr} (H^{6}) ] &= 5 n^2 \left(n^2+2\right)\\ \mathbb{E}[ \mathrm{tr} (H^{8}) ] &= 7 n \left(2 n^4+10 n^2+3\right)\\ \mathbb{E}[ \mathrm{tr} (H^{10}) ] &=84 n^2 \left(32 n^4+80 n^2+23\right)\\ \mathbb{E}[ \mathrm{tr} (H^{12}) ] &=66 n \left(256 n^6+1120 n^4+784 n^2+45\right)\\ \end{align*} Now one sees that the expectation values are invariant (up to sign) under the transformation \( n \mapsto -n \). With a formula \begin{equation} \mathbb{E}[ \mathrm{tr} (H^{2 p}) ]_{\mathrm{GUE}(n)} =(-1)^{p+1} \mathbb{E}[ \mathrm{tr} (H^{2 p}) ] _{\mathrm{GUE}(-n)} \end{equation} One could say that the Gaussian unitary ensemble is self-dual.

Further comments and links:

- A proof of this duality can be found in "Duality of real and quaternionic random matrices by Bryc and Pierce, 2008. I have not read their proof, but the paper seems well written and has lots of interesting references. Their paper proves an even more general duality than above, namely for all expectation values of products of matrix traces.
- Other papers on this duality that could be interesting are
- Mkrtchyan, “The Equivalence Of Sp(2N) and SO(-2N) Gauge Theory,” 1981
- On duality and negative dimensions in the theory of Lie groups and symmetric spaces, Mkrtchyan and Veselov,2010
- A detailed description of the Gaussian symplectic ensemble can be found in a previous post.
- I calculated the above expectation values with a Mathematica program. The program uses recursion based on Wick's theorem and some ideas from Derevianko's program Wick.m
- For all three ensembles I have normalized the probability density such that \begin{equation*} P(H) dH = \mathcal{N} dH \ \exp \left( -\frac{1}{2}\ \mathrm{tr}(H^2)\right) \end{equation*}
- The above expectation values are for matrices of size \( 2 n \times 2 n \) in the GSE and GOE, and matrices of size \( n \times n\) in the GUE.

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