Part (a): Joint distribution
The trace of \(H^2\) is:
\[
\mathrm{Tr}(H^2) = a^2 + b^2 + 2c^2.
\]
(The factor 2 on \(c^2\) arises because \(c\) appears both above and below the diagonal.) The GOE measure is:
\[
P(a, b, c) \propto \exp\!\left(-\frac{a^2 + b^2 + 2c^2}{2}\right).
\]
This means \(a, b \sim \mathcal{N}(0, 1)\) independently, and \(c \sim \mathcal{N}(0, 1/2)\) (the off-diagonal has half the variance).
Part (c): Deriving the Wigner surmise
Step 1: Change variables. Let \(t = a + b\), \(x = a - b\), \(y = 2c\). Then \(a = (t+x)/2\), \(b = (t-x)/2\), \(c = y/2\), and:
\[
a^2 + b^2 + 2c^2 = \frac{t^2 + x^2}{2} + \frac{y^2}{2} = \frac{t^2 + x^2 + y^2}{2}.
\]
The Jacobian is \(|\partial(a,b,c)/\partial(t,x,y)| = 1/4\). The distribution becomes:
\[
P(t, x, y) \propto \exp\!\left(-\frac{t^2}{4}\right) \cdot \exp\!\left(-\frac{x^2 + y^2}{4}\right).
\]
Step 2: Integrate out the trace. The variable \(t\) (the trace \(\mathrm{Tr}(H)\)) decouples completely and integrates to a constant \(\int e^{-t^2/4}\,dt = 2\sqrt{\pi}\).
Step 3: Switch to polar coordinates. The spacing is \(s = \sqrt{x^2 + y^2}\). In polar coordinates \((s, \theta)\) with \(x = s\cos\theta\), \(y = s\sin\theta\):
\[
dx\,dy = s\,ds\,d\theta.
\]
Integrating over the angle \(\theta \in [0, 2\pi)\) gives a factor \(2\pi\):
\[
\boxed{P(s) = \frac{s}{2}\,e^{-s^2/4}, \qquad s \geq 0.}
\]
(Normalization: \(\int_0^\infty \frac{s}{2} e^{-s^2/4}\,ds = 1\), verified by substitution \(u = s^2/4\).)
Physical interpretation: The factor \(s\) in \(P(s) \propto s\) as \(s \to 0\) is the linear level repulsion of the GOE (\(\beta = 1\)). Compare: the GUE has \(P(s) \propto s^2\) (quadratic; the same \(\beta = 2\) repulsion appears for the CUE in Exercise 2.3), and the Poisson ensemble (no correlations) has \(P(s) \propto \text{const}\) (no repulsion).