$$\DeclareMathOperator\Imm{Imm}$$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?

• Would you say what you denote by $\mathbb{H}_n^+$? – YCor Nov 8 at 15:06
• @YCor. The cone of positive semi-definite Hermitian matrices. – Denis Serre Nov 8 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 $$|\det C|^2=\det(C^TC)=\det(A)$$.
• 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. – Mark Wildon Nov 8 at 18:56
• 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". – Denis Serre Nov 8 at 19:51
• @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. – Suvrit Nov 9 at 1:27
• I have fixed the proof now (to use CS correctly!) and undeleted. – Suvrit Nov 9 at 4:26
• @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. – Denis Serre Nov 9 at 8:22