$\DeclareMathOperator\Imm{Imm}$I am looking for a proof in English or French of Schur's theorem that, for every $H$ in the space $\mathbb H_n^+$ of positive semi-definite Hermitian matrices, and every irreducible character $\chi$ of $\mathfrak S_n$, $\chi(e)\det H\le\Imm_\chi(H)$, where the immanant $\Imm_\chi$ is defined by $$\Imm_\chi(H):=\sum_\sigma\chi(\sigma)\prod_{i=1}^nh_{i\sigma(i)}.$$ Notice that the original paper I. Schur, "Über endlicher Gruppen und Hermiteschen Formen" Math. Z., 1 (1918) pp. 184–207, is in German.

By the way, it seems that many authors relate Schur's theorem to symmetric polynomials. Is there any purely representation-theoretic proof of the inequality above? Let $(\rho,V)$ be a unitary representation whose character is $\chi$. We may associate to $\Imm_\chi(H)$ a Hermitian matrix over $V$ by $$K_\rho:=\sum_\sigma\left(\prod_{i=1}^nh_{i\sigma(i)}\right)\rho(\sigma).$$ It would be sufficient to prove that $K\ge(\det H)I_V$, where $I_V$ denotes the matrix of the scalar product. Because of Frobenius's theorem about the orthogonal decomposition of the regular representation, this amounts to proving that the analogous sum, where $\rho$ is replaced by the regular representation, satisfies the same estimate. In other words, Schur's theorem would be implied by the inequality $$\forall \xi\in{\mathbb C}^{\frak S_n},\,\forall H\in{\mathbb H}_n^+,\qquad |\xi|^2\det H\le\sum_{\sigma,\theta}\bar\xi_\sigma\xi_\theta\prod_ih_{\sigma(i)\theta(i)}.$$ Is this inequality true?

  • 1
    $\begingroup$ Would you say what you denote by $\mathbb{H}_n^+$? $\endgroup$ – YCor Nov 8 '18 at 15:06
  • $\begingroup$ @YCor. The cone of positive semi-definite Hermitian matrices. $\endgroup$ – Denis Serre Nov 8 '18 at 15:47

Many thanks to Denis for pointing out my erroneous initial "proof". This time around the proof is correct, and directly proves the assertion in line 3 of the OP, i.e., $\chi(e)\det(A)\le d_\chi(A)$ (I will write $d_\chi(I)$ instead of $\chi(e)$ for uniformity).

The explicit notation is cumbersome, so I am just writing a proof sketch.

  1. First, recall that $d_\chi(A)=z^T(\otimes^n A)z$ for a suitable vector $z$
  2. Next, use Cauchy-Schwarz to obtain $$|z^T(\otimes^n (X^TY))z|^2 = |z^T(\otimes^n X^T)(\otimes^n Y)z|^2\le z^T(\otimes^n X^TX)z \cdot z^T(\otimes^n Y^TY)z$$
  3. Now write $A=C^TC$ for some upper triangular matrix $C$ (since $A$ is PSD we can do this). Then, put $X=C$ and $Y=I$ above, to obtain
  4. $|z^T(\otimes^n C)z|^2 = |z^T(\otimes^n I)z|^2|\det C|^2 \le |z^T(\otimes^n C^TC)z|\cdot |z^T(\otimes^n I)z|$, where we used the upper triangular nature of $C$ for the first step. In other words, we have shown that
  5. $d_\chi(I)^2 \det(A) \le d_\chi(A)d(I)$, since $|\det C|^2=\det(C^TC)=\det(A)$.
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    $\begingroup$ To get the immanent inequality from Schur's Theorem, take $x_\sigma = \chi(\sigma)$. The coefficient of $a_{1\rho(1)}\ldots a_{n\rho(n)}$ in the left-hand side is then $\sum_{\sigma, \tau : \tau\sigma^{-1} = \rho} \chi(\sigma) \chi(\tau) = \sum_{\sigma} \chi(\sigma)\chi(\rho\sigma) = \sum_{\sigma} \chi(\sigma^{-1})\chi(\rho\sigma) = |G| \chi(\rho) / \chi(1)$ by an orthogonality relation. So the left-hand side is $|G|/\chi(1)$ times the immanent sum, and the right-hand side is $|G| \mathrm{det}(A)$ again by character orthogonality. $\endgroup$ – Mark Wildon Nov 8 '18 at 18:56
  • $\begingroup$ I am dubious about the use of Cauchy-Schwarz. First, $B$ is not symmetric (or Hermitian). Second, you write an inequality whose sense is opposite to that of "Schur's Theorem". $\endgroup$ – Denis Serre Nov 8 '18 at 19:51
  • $\begingroup$ @DenisSerre indeed, you are right I flipped the inequalities, and the "proof" is incorrect as written. Time to dig into multilinear algebra to get a clean proof. Apologies for the rushed incorrect answer. Deleting it now. $\endgroup$ – Suvrit Nov 9 '18 at 1:27
  • $\begingroup$ I have fixed the proof now (to use CS correctly!) and undeleted. $\endgroup$ – Suvrit Nov 9 '18 at 4:26
  • $\begingroup$ @Suvrit. You should develop point 1, because otherwise, one does not see the role of the assumption that $\chi$ is a character. I see the definition of $z$, whose coordinate $z_{i_1\cdots i_n}$ vanishes unless $k\mapsto i_k$ is a permutation, in which case it equals $\chi(i)$. But still the formula depends upon a non trivial identity, valid only for characters. $\endgroup$ – Denis Serre Nov 9 '18 at 8:22

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