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I have a vector space which is a tensor product of two vector spaces, of dimensions $d_1, d_2$ respectively.

Consider Haar random unitaries acting on the full space with matrix elements $U_{i_1 j_1, i_2j_2}$, where indices $i_1,i_2$ run over the first space and $j_1,j_2$ run over the second space.

Define a new matrix with matrix elements $V_{i_1,j_1} := U_{i_1 j_1, 00}$. Is the distribution of $V$ well-known? (Perhaps only in the special cases $d_1=d_2$?)

Additionally, in the case of random unitaries Weingarten calculus tells us the expectation values of products of the matrix elements of $U$ (more precisely, occurring in a balanced polynomial). Is there a similar calculation for the matrix elements of $V$?

Note: $U_{i_1 j_1,00}$ defines a Haar random vector, and $V$ amounts to reshaping the legs of the vector into a matrix.

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Assuming real matrix elements, the distribution of the $N=d_1d_2$ elements $v_1,v_2,\ldots v_N$ of the $d_1\times d_2$ matrix $V$ is a delta function, $$P(v_1,v_2,\ldots v_N)\propto\delta\left(1-\sum_{j=1}^N v_j^2\right).$$ The marginal distribution of $k$ out of these $N$ elements is obtained by integrating out the other elements, $$P_{k}(v_{N-k+1},\ldots v_N)\propto\left(1-\sum_{j=N-k+1}^N v_j^2\right)^{(N-k)/2-1}\theta\left(1-\sum_{j=N-k+1}^n v_j^2\right),$$ with $\theta$ the unit step function. See this older MO post for a derivation.

For large $N\gg 1$ the matrix elements of $V$ are independent Gaussian with mean zero and variance $1/N$.


For complex matrix elements $v_j=x_j+iy_j$, the distribution of the $2N$ variables $\mathbf{x}=x_1,x_2,\ldots x_N$, $\mathbf{y}=y_1,y_2,\ldots y_N$ is given by $$P(\mathbf{x},\mathbf{y})\propto\delta\left(1-\sum_{j=1}^N (x_j^2+y_j^2)\right).$$ The marginal distribution becomes $$P_{k}(v_{N-k+1},\ldots v_N)\propto\left(1-\sum_{j=N-k+1}^N (x_j^2+y_j^2)\right)^{N-k-1}\theta\left(1-\sum_{j=N-k+1}^n (x_j^2+y_j^2)\right).$$

For large $N\gg 1$ the real and imaginary parts of the matrix elements of $V$ are independent Gaussian with mean zero and variance of $1/(2N)$.

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  • $\begingroup$ Thanks. The matrix elements are complex in general because U is a unitary. Does your answer change correspondingly? $\endgroup$
    – nervxxx
    Commented Sep 3, 2022 at 9:59
  • $\begingroup$ I added the complex case. $\endgroup$ Commented Sep 3, 2022 at 10:42

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