# Average of the maximum matrix element over the Haar measure

Let $$U$$ be a $$d\times d$$ unitary matrix, and $$U_{i,j}$$ be its matrix elements. I am interested in the following quantity $$\int dU \max_j |U_{1,j}|^2 \ ,$$ where $$dU$$ is the uniform Haar measure over $${\rm SU}(d)$$.

Please let me know if you have any idea for calculating this integral for general $$d$$.

• Your title asks for the maximum matrix element (out of, presumably, all of them), but your integral only selects the maximum element in the first row. Which do you mean? – LSpice Dec 15 '19 at 2:15
• I believe I've read that the joint distribution of the $|U_{1,j}|^2$ is the uniform distribution on the standard $d$-simplex. – MTyson Dec 15 '19 at 4:11

$$\int dU \max_j |U_{1,j}|^2 =\frac{1}{d}\sum_{j=1}^d \frac{1}{j}.$$
For large $$d$$ this tends to $$(1/d)\log d$$. The complete probability distribution of the row-maximum is known.
The maximum of all matrix elements is more difficult and only large-$$d$$ asymptotics has a closed-form expression, see Maxima of entries of Haar distributed matrices (alternative link).
If $$W_d$$ is the maximum matrix element in absolute value of a Haar-distributed $$d\times d$$ unitary matrix, then $$W_d^2\rightarrow (2/d)\log d$$ in probability for $$d\rightarrow\infty$$, so twice as large as for the row-maximum.
$$\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}$$Any linear isometry of $$\C^d$$ is a unitary transformation. Therefore, the distribution of the random vector $$V:=(X_1,Y_1,\dots,X_d,Y_d)$$ is uniform on the unit sphere in $$\R^{2d}$$, where $$X_j:=\Re U_{1,j}$$ and $$Y_j:=\Im U_{1,j}$$. So, $$V$$ equals $$W:=(W_1,\dots,W_{2d})$$ in distribution, where $$W_j:=\frac{Z_j}{\sqrt{Z_1^2+\dots+Z_{2d}^2}}$$ and $$Z_1,\dots,Z_{2d}$$ are iid standard normal random variables (r.v.'s). So, the random vector $$(|U_{1,1}|^2 ,\dots,|U_{1,d}|^2)$$ equals $$R:=(R_1,\dots,R_d)$$ in distribution, where $$R_j:=\frac{T_j}{T_1+\dots+T_d}$$ and $$T_k:=Z_{2k-1}^2+Z_{2k}^2$$, so that $$T_1,\dots,T_d$$ are iid standard exponential r.v.'s. So, $$\max_j|U_{1,j}|^2$$ equals $$\max_j R_j$$ in distribution. The distribution and, in particular, the expectation of $$\max_j R_j$$ were found a long time ago; see e.g. Irwin (1955) and historical references therein going back to as far as 1897. In particular, according to formula (15) in Cochran's paper, $$E\max_j|U_{1,j}|^2=E\max_j R_j=\frac1d\,\Big(1+\frac12+\dots+\frac1d\Big).$$