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1. If $p>q$: then $r>q$ and after pairwise elimination of $q$ $\epsilon$-$\tau$ pairs, one is left with $q$ direct contractions of vectors and covectors mediated by a permutation in $S_q$ (this is the Schur-Weyl or $GL$ invariant part) as well as $(p-q)/d$ surviving $\tau$'s giving what the classics called "Klammerfaktoren" involving vectors only.
2. If $p<q$: then $r<q$ and after pairwise elimination of $r$ $\epsilon$-$\tau$ pairs, one is left with $p$ direct contractions of vectors and covectors mediated by a permutation in $S_p$ (this is again the Schur-Weyl or $GL$ invariant part) as well as $(q-p)/d$ surviving $\epsilon$'s giving "Klammerfaktoren" involving covectors only.
3. What you are interested in is the special case $p=q=n$, or pure Schur-Weyl situation with no "Klammerfaktoren".

My reply: The Haar measure is just salad dressing, so the structure of the proof becomes natural. There are three steps only: 1) average over the group. 2) figure out what averaging means in algebro/combinatorial terms rather than analytic ones 3) do the calculation and contemplate the result. It would have taken me too long to draw pictures in my answer but this is a purely graphical proof. If you want to see these pictures at least in the case $d=2$, $p$ arbitrary and $q=0$ case, then see pages 16-17 of this article. (If you have access to JKTR, the published version is quite a bit better).

I could have written my first answer in a purely combinatorial way without mentioning the Haar measure at all but then some elements of the proof would have come totally out the blue. In particular, I would have had to all of a sudden say: it $T$ is invariant then $$ T=c_n\ ({\rm det}(\partial g))^n\ gT $$ where $gT$ is the expression obtained after Cramer's rule and the $g$'s are treated as formal variables. So again I insist, my answer is purely combinatorial. But I don't know if it may be useful in nonzero characteristic. This may require understanding the arithmetic of the $c_n$'s.

Edit: Yet another rewriting of the proof in order to eliminate analysis and Haar measures etc.

Let me simply say that this is not my proof but Clebsch's in his amazing article "Ueber symbolische Darstellung algebraischer Formen" in Crelle 1861. (Please click on the full text link and read Section 3 of that article and in particular pages 12 and 13 which contain the sum of squares non-vanishing argument).

1. If $p>q$: then $r>q$ and after pairwise elimination of $q$ $\epsilon$-$\tau$ pairs, one is left with $q$ direct contractions of vectors and covectors mediated by a permutation in $S_q$ (this is the Schur-Weyl or $GL$ invariant part) as well as $(p-q)/d$ surviving $\tau$'s giving what the classics called "Klammerfaktoren" involving vectors only.
2. If $p<q$: then $r<q$ and after pairwise elimination of $r$ $\epsilon$-$\tau$ pairs, one is left with $p$ direct contractions of vectors and covectors mediated by a permutation in $S_p$ (this is again the Schur-Weyl or $GL$ invariant part) as well as $(q-p)/d$ surviving $\epsilon$'s giving "Klammerfaktoren" involving covectors only.
3. What you are interested in is the special case $p=q=n$, or pure Schur-Weyl situation with no "Klammerfaktoren". In other words, Schur-Weyl duality part b) is this particular case of the First Fundamental theorem of classical invariant theory for $SU(d)$ slash $SL(d)$.

My reply: The Haar measure is just salad dressing. I incorporated into the narrative in order to make the structure of the proof more natural. The proof I gave for the FFT of $SU(d)$ slash $SL(d)$ has three easy steps only: 1) average over the group. 2) figure out what averaging means in algebro/combinatorial terms rather than analytic ones 3) do the calculation and contemplate the result. It would have taken me too long to draw pictures in my answer but this is a purely graphical proof. If you want to see these pictures at least in the case $d=2$, $p$ arbitrary and $q=0$ case, then see pages 16-17 of this article. (If you have access to JKTR, the published version is quite a bit better).

I could have written my first answer in a purely combinatorial way without mentioning the Haar measure at all but then some elements of the proof would have come totally out the blue. In particular, I would have had to all of a sudden say: if $T$ is invariant then $$ T=c_n\ ({\rm det}(\partial g))^n\ gT $$ where $gT$ is the expression obtained after Cramer's rule and the $g$'s are treated as formal variables. So again I insist, my answer is purely combinatorial. But I don't know if it may be useful in nonzero characteristic. This may require understanding the arithmetic of the $c_n$'s.

Edit: Yet another rewriting of the proof of the FFT for $SU(d)$ slash $SL(d)$ in order to eliminate analysis and Haar measures etc.

Let me simply say that this is not my proof but Clebsch's in his amazing article "Ueber symbolische Darstellung algebraischer Formen" in Crelle 1861. (Please click on the full text link and read Section 3 of that article and in particular pages 12 and 13 which contain the sum of squares non-vanishing argument). As another comment about history, I put quotes when talking about the "Cayley identity" because (of course Arnold would say) it is nowhere to be found in the works of Cayley. The earliest instance I have seen is in Clebsch's book on binary forms for $d=2$. No doubt, he must have been trying to get a better understanding of the $\rho_n$ coefficients and also the Gordan-Clebsch series (see my JKTR article). Tony Crilly, Alan Sokal and I are supposed to work on an article on the history of the "Cayley identity" but we have been distracted by other tasks. It is on the $({\rm back})^n$-burner with $n$ large.

My reply: The Haar measure is just salad dressing, so the structure of the proof becomes natural. There three steps only: 1) average over the group. 2) figure out what averaging means in algebro/combinatorial terms rather than analytic ones 3) do the calculation and contemplate the result. It would have taken me too long to draw pictures in my answer but this is a purely graphical proof. If you want to see these pictures at least in the case $d=2$, $p$ arbitrary and $q=0$ case, then see pages 16-17 of this article. (If you have access to JKTR, the published version is quite a bit better).

Let me simply say that this is not my proof but Clebsch's in his amazing article "Ueber symbolische Darstellung algebraischer Formen" in Crelle 1861. (Please click on the full text link and read Section 3 of that article).

My reply: The Haar measure is just salad dressing, so the structure of the proof becomes natural. There are three steps only: 1) average over the group. 2) figure out what averaging means in algebro/combinatorial terms rather than analytic ones 3) do the calculation and contemplate the result. It would have taken me too long to draw pictures in my answer but this is a purely graphical proof. If you want to see these pictures at least in the case $d=2$, $p$ arbitrary and $q=0$ case, then see pages 16-17 of this article. (If you have access to JKTR, the published version is quite a bit better).

Let me simply say that this is not my proof but Clebsch's in his amazing article "Ueber symbolische Darstellung algebraischer Formen" in Crelle 1861. (Please click on the full text link and read Section 3 of that article and in particular pages 12 and 13 which contain the sum of squares non-vanishing argument).

Let us now assume that $q=0$ and that $T\in\mathcal{T}_{p,0}$ is invariant, i.e., $gT=T$ for all $g\in SU(d)$ or rather $SL(d)$. Since any $g\in GL(d)$ can be written as $g=\lambda h$ with $h\in SL(d)$ and $\lambda$ some $d$-th root of ${\rm det}(g)$ then since $gT$ is a homogeneous polynomial we more generally have $gT=({\rm det}(g))^n\ T$ for all $g\in GL(d)$, and even for all $g\in {\rm Mat}_{d\times d}$. Here $n$ denotes the weight $p/d$. We now have $$ ({\rm det}(\partial g))^n\ gT=({\rm det}(\partial g))^n\ ({\rm det}(g))^n=\rho_n\ T $$ The heart of the proof (as well as of the understanding of the Haar measure as pure combinatorics), is to show that $$ \rho_n=({\rm det}(\partial g))^n\ ({\rm det}(g))^n \neq 0 $$ The most precise way of doing this is via the "Cayley identity" exercise which gives the explicit formula $$ \rho_n=\frac{1}{c_n} $$ But a cheaper argument goes as follows. When summing over the permutation in $S_{nd}$ which I called a Wick contraction scheme (just the Leibnitz rule) arising in the computation of $\rho_n$, one gets a sum of squares: what I called the two decoupled (rather than connected) components that I mentioned earlier are identical and deliver the same numerical evaluation. One just need choose one of these contraction schemes, for instance, coming from $$ \left[({\rm det}(\partial g))\ ({\rm det}(g))\right]^n $$ and check it is nonzero. The germ of this "mirror symmetry" between the two components is visible in the right-hand side of the basic identity I used $$ \frac{\partial}{\partial g_{cl}}g_{i b}=\delta_{ci}\delta_{lb} $$

Let us now assume that $q=0$ and that $T\in\mathcal{T}_{p,0}$ is invariant, i.e., $gT=T$ for all $g\in SU(d)$ or rather $SL(d)$. Since any $g\in GL(d)$ can be written as $g=\lambda h$ with $h\in SL(d)$ and $\lambda$ some $d$-th root of ${\rm det}(g)$ then since $gT$ is a homogeneous polynomial we more generally have $gT=({\rm det}(g))^n\ T$ for all $g\in GL(d)$, and even for all $g\in {\rm Mat}_{d\times d}$. Here $n$ denotes the weight $p/d$. We now have $$ ({\rm det}(\partial g))^n\ gT=({\rm det}(\partial g))^n\ ({\rm det}(g))^n T=\rho_n\ T $$ The heart of the proof (as well as of the understanding of the Haar measure as pure combinatorics), is to show that $$ \rho_n=({\rm det}(\partial g))^n\ ({\rm det}(g))^n \neq 0 $$ The most precise way of doing this is via the "Cayley identity" exercise which gives the explicit formula $$ \rho_n=\frac{1}{c_n} $$ But a cheaper argument goes as follows. When summing over the permutation in $S_{nd}$ which I called a Wick contraction scheme (just the Leibnitz rule) arising in the computation of $\rho_n$, one gets a sum of squares: what I called the two decoupled (rather than connected) components that I mentioned earlier are identical and deliver the same numerical evaluation. One just need choose one of these contraction schemes, for instance, coming from $$ \left[({\rm det}(\partial g))\ ({\rm det}(g))\right]^n $$ and check it is nonzero. The germ of this "mirror symmetry" between the two components is visible in the right-hand side of the basic identity I used $$ \frac{\partial}{\partial g_{cl}}g_{i b}=\delta_{ci}\delta_{lb} $$