On a proof involving Young symmetrizers acting on tensor spaces I hope this is not too elementary for this site, but I already asked something similar on MSE which has not received any attention whatsoever. I am extremely unfamiliar with the algebraic/representation theoretic area where Young tableaux are frequently used, so I am posting it here slightly rewritten as compared to the MSE question hoping someone can help me understand this without me having to learn an entire area from skratch just for this.

My motivation is that I want to understand a proof given by Krupka in a paper (page 77 at formula (8) ) regarding total divergence equations in the calculus of variations.
If I remove the algebraic problem from its context, there is a finite $m$ dimensional real vector space $V$ and a tensor $T\in \left(S^r V\right)^{\otimes m}\otimes V\subseteq V^{\otimes mr+1}$ given in index notation as $$ T^{I_1...I_mj}, $$with each multiindex $I_k=(i_{1,(k)},\dots,i_{r,(k)})$ standing for a block of $r$ indices in which $T$ is symmetric.
This tensor observes the additional symmetry $$ T^{I_1...(I_k|...I_m|j)}=0\qquad(\sharp), $$i.e. when symmetrized over $I_kj$ for any $k=1,2,\dots,m$, it gives zero.
Krupka now proves that $$ T^{I_1...I_mj}=0\qquad(\ast). $$
As the proof technique Krupka gives, verbatim

We prove $(\ast)$ by showing that all Young diagrams, defining the Young decomposition of the expression on the left of $(\ast)$, vanish.

I already have some trouble interpreting this sentence, but I'll ask about it later.
Then the proof proceeds as follows:

*

*Since $T$ is symmetric in each of the blocks $I_k$, only those Young tableaux contribute in which every element of each block $I_k$ lies on one row.

*If $j$ is in the same row as one of the $I_k$, then by $(\sharp)$, the tableau gives zero. Thus $j$ goes to the last box of the tableau.

*If any row contains at least two multiindices $I_k$ then by taking column permutations, it is possible to bring $j$ to be in one row with an intact multiindex $I_k$, and we again get zero by 2.

*Thus the only possible nonzero tableau is $$\begin{matrix} I_1 \\ I_2 \\ \vdots \\ I_m \\ j\end{matrix}, $$ but the first column of this tableau has $m+1$ boxes to antisymmetrize on and as $\dim V=m$, hence this also gives zero. Therefore $T=0$.


Due to my unfamiliarity with Young symmetrizers, Schur modules etc. although I understand the general thought behind this proof, I am having trouble getting a precise and complete understanding.

*

*Just the general method of attack. From that I know the relevant tensor spaces can be decomposed as $$ \left(S^rV\right)^{\otimes m}\otimes V=\bigoplus_\lambda S^\lambda(V), $$where the $\lambda$ are numbered Young tableaux with $mr+1$ boxes and the $S^\lambda(V)$ are the corresponding Schur modules. The sum is over an appropriate set of Young tableaux. $$\ $$As far as I am aware all tableaux of the same shape give isomorphic Schur modules, and this sum has multiplicities. But tableaux with the same shape but different numbering give subspaces of $V^{\otimes mr+1}$ that are not literally the same as point sets. $$\ $$If $c_\lambda:V^{\otimes mr+1}\rightarrow S^\lambda(V)$ is the corresponding Young symmetrizer (symmetrizes the rows first, then antisymmetrizes the columns), then a scalar multiple of $c_\lambda$ is a projection, hence we should have $$ T=\sum_\lambda n_\lambda c_\lambda T. $$ So I suppose the idea behind the proof is to show that all Young symmetrizers $c_\lambda$ give zero when applied to $T$ right?

*The first step 1. in the list above is somewhat problematic for me. At first I assumed that if there is a tensor symmetric in a number of indices, then any Young symmetrizer is zero on it whenever the symmetric indices are not in the same row of the tableau. But this is manifestly false, for example if the tensor $\phi^{ijk}$ is symmetric in $ij$ but the symmetrizer corresponding to the tableau $$\begin{matrix} i & k \\ j\end{matrix}$$ gives (up to a scalar multiple) $$ (c_\lambda\phi)^{ijk}=\phi^{kji}-\phi^{kij}, $$ which is nonzero in general. So why can we disregard every tableau in which a multiindex $I_k$ is spread over multiple rows?

If my concerns above are answered then I understand the rest of the proof. So my questions are as above. However basically any answer that helps me understand Krupka's proof is helpful to me.
One thing I'd like to remark is that with the first point 1. it would be helpful to use Young symmetrizers with the alternative convention, i.e. antisymmetrize first and then symmetrize last. However then Step 2. would be ruined. Also Krupka later on (page 79 at formula (11) ) deduces that an object is antisymmetric in a set of indices from a similar argument except where Step 4. does not give a zero symmetrizer. Hence, Krupka certainly uses the "antisymmetrize last" convention.

Edit: As mentioned in the comments, I'd prefer to learn how this proof works as opposed to providing some alternative proof for the same result, since I wish to explore some of its implications and also generalize it.
One thing I thought about is that there is a rule $$ S_\lambda V\otimes S^rV=\bigoplus_\mu S_\mu V $$where the $\mu$ are tableaux obtained from $\lambda$ by adding $r$ boxes such that no two added boxes are in the same column.
Hence we can exclude those tableaux in which different elements of $I_k$ are in the same column. I then thought we can make column permutations to only consider tableaux in which each $I_k$ is on one row.
But this is also false, at least without some further arguments, because one cannot make column permutations in a tableaux freely without changing the Young symmetrizer since $c_{\pi\cdot\lambda}=\pi\cdot c_\lambda\cdot\pi^{-1}$, and if $\pi$ is a column permutation then $\pi\cdot c_\lambda=\pm c_\lambda$ but the additional factor of $\pi^{-1}$ on the right still remains.
I thought the answer was something simple I missed, but I strongly think now Krupka's proof is flawed or at least incomplete. The results he derives in the linked paper are known to be correct however (there are alternative proofs), so I suspect there is a way to make this proof sound. I am still quite interested in that.
 A: I don't know what the paper is doing -- its use of Young diagrams definitely
looks vague to me, but maybe I would understand it better after reading (e.g.)
Towber. But here is an approach that avoids any combinatorics.
I generalize the problem: First, $\mathbb{R}$ becomes any field $K$ (not sure
if fieldness is really necessary); second, I replace the symmetric power
$S^{r}V$ by the whole symmetric algebra $SV$, of which the symmetric power
$S^{r}V$ is just a direct addend. Thus, the situation is as follows:
We are given a field $K$ and an $m$-dimensional $K$-vector space $V$. Let $SV$
denote the symmetric algebra of the vector space $V$.
We let $\left[  m\right]  :=\left\{  1,2,\ldots,m\right\}  $. For each
$i\in\left[  m\right]  $, we let $\mu_{i}:\left(  SV\right)  ^{\otimes
m}\otimes V\rightarrow\left(  SV\right)  ^{\otimes m}$ be the $K$-linear map
that sends each pure tensor $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{m}\otimes
b\in\left(  SV\right)  ^{\otimes m}\otimes V$ to
\begin{align*}
a_{1}\otimes a_{2}\otimes\cdots\otimes a_{i-1}\otimes a_{i}b\otimes
a_{i+1}\otimes a_{i+2}\otimes\cdots\otimes b
\end{align*}
(where the product $a_{i}b$ is taken in the symmetric algebra $SV$). This map
$\mu_{i}$ can be viewed as a "partial symmetrization": It multiplies the $b$
onto the $a_{i}$ in $SV$ (which, if $\operatorname*{char}K=0$ and if you view
$SV$ as a subspace of the tensor algebra $TV$, corresponds to symmetrizing the
tensor $a_{i}\otimes b$).
Now, you claim:

Theorem 1. We have $\bigcap_{i\in\left[  m\right]  }\operatorname*{Ker}
\mu_{i}=0$.

Proof. It is well-known that we can identify the symmetric algebra $SV$ with
a polynomial ring $K\left[  u_{1},u_{2},\ldots,u_{m}\right]  $ in $m$
variables. To do so, we fix a basis $\left(  e_{1},e_{2},\ldots,e_{m}\right)
$ of $V$; then, there is a unique $K$-algebra isomorphism
\begin{align*}
f:K\left[  u_{1},u_{2},\ldots,u_{m}\right]    & \rightarrow SV,\\
u_{i}  & \mapsto e_{i}.
\end{align*}
Consider this $f$. Let $K\left[  u_{1},u_{2},\ldots,u_{m}\right]
_{\operatorname*{lin}}$ denote the degree-$1$ homogeneous component of
$K\left[  u_{1},u_{2},\ldots,u_{m}\right]  $. Then, $f\left(  K\left[
u_{1},u_{2},\ldots,u_{m}\right]  _{\operatorname*{lin}}\right)  =S^{1}V$ (the
$1$-st symmetric power of $V$). We identify $S^{1}V$ with $V$.
Thus, via the isomorphism $f^{\otimes m}$, we can identify the tensor power
$\left(  SV\right)  ^{\otimes m}$ with
\begin{align*}
\left(  K\left[  u_{1},u_{2},\ldots,u_{m}\right]  \right)  ^{\otimes m}  &
\cong\bigotimes_{i=1}^{m}\left(  K\left[  u_{i,1},u_{i,2},\ldots
,u_{i,m}\right]  \right)  \\
& \cong K\left[  u_{1,1},u_{1,2},\ldots,u_{m,m}\right]
\end{align*}
(a polynomial ring in $m^{2}$ indeterminates, named $u_{i,j}$ for all pairs
$\left(  i,j\right)  \in\left\{  1,2,\ldots,m\right\}  ^{2}$). Likewise, via
the isomorphism $f^{\otimes\left(  m+1\right)  }$, we can identify our tensor
product $\left(  SV\right)  ^{\otimes m}\otimes V$ with
\begin{align*}
& \left(  K\left[  u_{1},u_{2},\ldots,u_{m}\right]  \right)  ^{\otimes
m}\otimes K\left[  u_{1},u_{2},\ldots,u_{m}\right]  _{\operatorname*{lin}}\\
& \cong\bigotimes_{i=1}^{m}\left(  K\left[  u_{i,1},u_{i,2},\ldots
,u_{i,m}\right]  \right)  \otimes K\left[  x_{1},x_{2},\ldots,x_{m}\right]
_{\operatorname*{lin}}\\
& \cong\left\{  \text{all polynomials in the }m^{2}+m\text{ indeterminates}
\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1}
,x_{2},\ldots,x_{m}\text{ that are}\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \text{homogeneous of degree }1\text{ in the last
}m\right.  \\
& \ \ \ \ \ \ \ \ \ \ \left.  \text{indeterminates }x_{1},x_{2},\ldots
,x_{m}\right\}  .
\end{align*}
For each $i\in\left[  m\right]  $, the map $\mu_{i}:\left(  SV\right)
^{\otimes m}\otimes V\rightarrow\left(  SV\right)  ^{\otimes m}$ corresponds
(under these two identifications) to the map that takes a polynomial $P$ in
the $m^{2}+m$ indeterminates $u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1}
,x_{2},\ldots,x_{m}$ and replaces the last $m$ indeterminates $x_{1}
,x_{2},\ldots,x_{m}$ by $u_{i,1},u_{i,2},\ldots,u_{i,m}$. (Check this -- it's
particularly easy here since $\mu_{i}$ is a $K$-algebra homomorphism when
extended to $\left(  SV\right)  ^{\otimes\left(  m+1\right)  }$ in the obvious
way.) We denote the image of $P$ under the latter map by $P\left(
\mathbf{x}\mapsto\mathbf{u}_{i}\right)  $.
Thus, $\bigcap_{i\in\left[  m\right]  }\operatorname*{Ker}\mu_{i}$ corresponds
to the set of all polynomials $P$ in the $m^{2}+m$ indeterminates
$u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1},x_{2},\ldots,x_{m}$ that are homogeneous
of degree $1$ in the last $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$ and
satisfy
\begin{align*}
P\left(  \mathbf{x}\mapsto\mathbf{u}_{1}\right)  =P\left(  \mathbf{x}
\mapsto\mathbf{u}_{2}\right)  =\cdots=P\left(  \mathbf{x}\mapsto\mathbf{u}
_{m}\right)  =0.
\end{align*}
Our goal is thus to show that the only such polynomial $P$ is $0$.
This is easiest to achieve by another change of viewpoint: We let $S$ be the
commutative ring $K\left[  u_{1,1},u_{1,2},\ldots,u_{m,m}\right]  $, and we
let $Q$ be the field of fractions of $S$. Any polynomial $P$ in the  $m^{2}+m$
indeterminates $u_{1,1},u_{1,2},\ldots,u_{m,m},x_{1},x_{2},\ldots,x_{m}$ over
our original field $K$ can then be viewed as a polynomial in the $m$
indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $S$, thus also as a polynomial
in the $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $Q$ (since $S$ is a
subring of $Q$). Moreover, if this polynomial $P$ is homogeneous of degree $1$
in the last $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$, then viewing it as
polynomial in the $m$ indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $Q$
yields a degree-$1$ homogeneous polynomial in these $m$ indeterminates, i.e.,
a linear form on the $m$-dimensional $Q$-vector space $Q^{m}$. Finally, the
condition
\begin{align*}
P\left(  \mathbf{x}\mapsto\mathbf{u}_{1}\right)  =P\left(  \mathbf{x}
\mapsto\mathbf{u}_{2}\right)  =\cdots=P\left(  \mathbf{x}\mapsto\mathbf{u}
_{m}\right)  =0
\end{align*}
tells us that the values of this linear form on the $m$ vectors $\left(
\begin{array}
[c]{c}
u_{1,1}\\
u_{1,2}\\
\vdots\\
u_{1,m}
\end{array}
\right)  $, $\left(
\begin{array}
[c]{c}
u_{2,1}\\
u_{2,2}\\
\vdots\\
u_{2,m}
\end{array}
\right)  $, $\ldots$, $\left(
\begin{array}
[c]{c}
u_{m,1}\\
u_{m,2}\\
\vdots\\
u_{m,m}
\end{array}
\right)  $ are all $0$. Since these $m$ vectors are $Q$-linearly independent
(because the matrix they form has determinant $\det\left(  \left(
u_{i,j}\right)  _{i,j\in\left[  m\right]  }\right)  \neq0$), this entails that
the linear form must be $0$. Hence, $P$ must be $0$. This proves Theorem 1.
$\blacksquare$
