Errata for Fulton's "Young tableaux"

Question

Fulton's Young tableaux is one of the best texts on the subject, one which I often recommend and cite for reference. Unlike Fulton/Lang and Fulton/Harris, it is neither an early-dawn draft nor a chatty informal introduction but aims (and often succeeds) at giving a streamlined and readable presentation of the subject. All the more I find it important to spot and correct the occasional errors in it.

I used to think it had none, until I learned of three seriously broken proofs in rapid succession:

Serious errors:

1. page 97, proof of Lemma 5 (§ 7.4): The number \begin{align*} n_{\lambda} & =\left\vert \left\{ \left( p_{1},q_{1},p_{2},q_{2}\right) \ :\ p_{i}\in R\left( T\right) ,\ q_{i}\in C\left( T\right) \right. \right. \\ & \ \ \ \ \ \ \ \ \ \ \left. \left. p_{1}q_{1}p_{2}q_{2}=1, \ \operatorname*{sgn}\left( q_{1}\right) =\operatorname*{sgn}\left( q_{2}\right) \right\} \right\vert \end{align*} (where $R\left( T\right) $ is the row-fixing group and $C\left( T\right) $ the column-fixing group of a given standard tableau $T$) is wrong, as it fails to satisfy the equation $\left( b_{T}\cdot a_{T}\right) \cdot v_{T} =n_{\lambda}\cdot v_{T}$. (It would satisfy the equation if it was true that any permutations $p_{1},p_{2}\in R\left( T\right) $ and $q_{1},q_{2}\in C\left( T\right) $ satisfying $p_{1}q_{1}p_{2}q_{2}=1$ would satisfy $\operatorname*{sgn}\left( q_{1}\right) =\operatorname*{sgn}\left( q_{2}\right) $. But this fails when $n=8$ and $\lambda = \left( 3,3,2\right) $ and $T= \begin{array} [c]{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & \end{array} $ and $p_{1}=\left[ 23164578\right] $ and $q_{1}=\left[ 48372615\right] $ and $p_{2}=\left[ 32146587\right] $ and $q_{2}=\left[ 18675342\right] $, where all permutations are given in one-line notation. Maybe it holds for two-row partitions? I don't know.)

Unfortunately, this is not easy to fix. We can make the equations $\left( b_{T}\cdot a_{T}\right) \cdot v_{T}=n_{\lambda}\cdot v_{T}$ and $\left( a_{T}\cdot b_{T}\right) \cdot\widetilde{v}_{T}=n_{\lambda}\cdot \widetilde{v}_{T}$ true if we replace the number $n_{\lambda}$ by \begin{align*} n_{\lambda}^{\prime}:=\sum_{\substack{\left( p_{1},q_{1},p_{2},q_{2}\right) ;\\p_{i}\in R\left( T\right) ,\ q_{i}\in C\left( T\right) ;\\ p_1q_{1}p_{2}q_{2}=1}}\operatorname*{sgn}\left( q_{1}\right) \cdot \operatorname*{sgn}\left( q_{2}\right) . \end{align*} But we also need to show that this number $n_{\lambda}^{\prime}$ is positive, and this is not as obvious as it was for $n_{\lambda}$. Fortunately, this can still be shown using some easily accessible results: The number $n_{\lambda }^{\prime}$ is easily seen to be coefficient of the identity permutation $1$ in $\left( a_{T}b_{T}\right) ^{2}=c_{T}^{2}$, where $c_T := a_T b_T$ (this differs a bit from Fulton's $c_T$, which is $b_T a_T$, but this is not a major distinction). However, it is known (see, e.g., Lemma 5.13.3 in Pavel Etingof et al., Introduction to representation theory, AMS 2011) that $c_{T}^{2}=\dfrac{n!}{\left\vert P_{\lambda}\right\vert \left\vert Q_{\lambda }\right\vert \dim V_{\lambda}}c_{T}$ for a certain positive number $\dfrac {n!}{\left\vert P_{\lambda}\right\vert \left\vert Q_{\lambda}\right\vert \dim V_{\lambda}}$ (see loc. cit. for its definition). (Note that the lemma is stated not for an arbitrary numbering $T$ but only for the specific numbering $T_{\lambda}$ which has the numbers $1,2,\ldots,n$ filled in from top to bottom, row by row. But this is fine, since the numbering can be chosen arbitrarily.) Furthermore, it is easy to see that the coefficient of the permutation $1$ in $c_{T}$ is $1$. Hence, the coefficient of $1$ in $c_{T}^{2}=\dfrac{n!}{\left\vert P_{\lambda}\right\vert \left\vert Q_{\lambda }\right\vert \dim V_{\lambda}}c_{T}$ is $\dfrac{n!}{\left\vert P_{\lambda }\right\vert \left\vert Q_{\lambda}\right\vert \dim V_{\lambda}}$, which is positive. Thus, $n_{\lambda}^{\prime}$ is positive, qed.

2. page 108, proof of Lemma 2 (§ 8.1): I am not convinced that "In the latter case, fixing $v_{p}=w_{1}$, it suffices to show that the difference of the two sides is an alternating function of $v_{1},v_{2},\ldots,v_{p},w_{2}$". The problem is that the left hand side is not linear in $v_{p}=w_{1}$, since $v_{p}$ appears twice in it (once as $v_{p}$ and once again as $w_{1}$). I don't know how easy this is to correct, but there are other proofs of Lemma 2 around in the literature (e.g., it is Exercise 6.66 in arXiv:2008.09862v3, where it is proved just using multi-column Laplace expansion).

3. page 113, proof of Lemma 4 (§ 8.2): The claim that "$e_{T^{\prime}}$ occurs in $g\cdot e_{T}$ with coefficient $1$" is not obvious. More importantly, this argument only rules out other highest weight vectors of the form $e_{T}$, but not highest weight vectors that are arbitrary linear combinations of different $e_{T}$'s. I don't know if this gap can be easily fixed. See an answer below for a correct (I hope...) proof of Lemma 4.

4. page 118, proof of Proposition 1 (§ 8.3): This proof makes a leap of faith in the last paragraph: How exactly does the construction of the map from $E^{\lambda}$ to $E\left( S^{\lambda}\right) $ yield that $E^{\lambda}\cong E\left( \widetilde{M}^{\lambda}\right) /E\left( Q^{\lambda}\right) $ ? There is no direct way to obtain this from the map itself, since it is not obvious that the map is injective. Instead, I suggest the following argument:

Proposition 4 in § 7.4 yields that $S^{\lambda}\cong\widetilde{M}^{\lambda }/Q^{\lambda}$, where $Q^{\lambda}$ is as defined in § 7.4. On the other hand, set $\mu=\widetilde{\lambda}$ and $\ell=\lambda_{1}$. Then, (11) shows that there is an isomorphism \begin{align*} H:E\left( \widetilde{M}^{\lambda}\right) \rightarrow\wedge^{\mu_{1}}\left( E\right) \otimes\cdots\otimes\wedge^{\mu_{\ell}}\left( E\right) \end{align*} (which sends any pure tensor $\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) \otimes_{\mathbb{C}\left[ S_{n}\right] }\left[ T\right] \in E\left( M^{\lambda}\right) $ to \begin{align*} \left( v_{T\left( 1,1\right) }\wedge v_{T\left( 1,2\right) }\wedge \cdots\wedge v_{T\left( 1,\mu_{1}\right) }\right) \otimes\cdots \otimes\left( v_{T\left( \ell,1\right) }\wedge v_{T\left( \ell,2\right) }\wedge\cdots\wedge v_{T\left( \ell,\mu_{\ell}\right) }\right) \end{align*} ).

However, the inclusion $Q^{\lambda}\rightarrow\widetilde{M}^{\lambda}$ yields (by the functoriality of $E$) a map from $E\left( Q^{\lambda}\right) $ to $E\left( \widetilde{M}^{\lambda}\right) $. Due to how $Q^{\lambda}$ was defined, the image of this map is the span of all the differences $\mathbf{v}\left( T\right) \otimes v_{T}-\sum_{S}\mathbf{v}\left( S\right) \otimes v_{S}$, where $\mathbf{v}$ ranges over all pure tensors in $E^{\otimes n}$, where $T$ varies over all numberings of $\lambda$, where $j$ and $k$ vary over all integers satisfying $1\leq j\leq\ell-1$ and $1\leq k\leq\mu_{j+1}$, and where the sum is over all numberings $S$ in $\pi_{j,k}\left( T\right) $. Under our above isomorphism $H$, these differences $\mathbf{v}\left( T\right) \otimes v_{T}-\sum_{S}\mathbf{v}\left( S\right) \otimes v_{S}$ turn into exactly the generators $\wedge\mathbf{v}-\sum\wedge\mathbf{w}$ of the subspace $Q^{\lambda}\left( E\right) $ defined in § 8.1. Hence, under the isomorphism $H$, the image of $E\left( Q^{\lambda}\right) $ in $E\left( \widetilde{M}^{\lambda}\right) $ becomes the subspace $Q^{\lambda}\left( E\right) $. Therefore, \begin{align*} E\left( \widetilde{M}^{\lambda}\right) /E\left( Q^{\lambda}\right) \cong\left( \wedge^{\mu_{1}}\left( E\right) \otimes\cdots\otimes\wedge ^{\mu_{\ell}}\left( E\right) \right) /Q^{\lambda}\left( E\right) =E^{\lambda} \end{align*} (by (4)).

But the functor $E$ is right-exact (since it is a tensor functor), and thus we have $E\left( \widetilde{M}^{\lambda}/Q^{\lambda}\right) \cong E\left( \widetilde{M}^{\lambda}\right) /E\left( Q^{\lambda}\right) \cong E^{\lambda}$. In other words, $E\left( S^{\lambda}\right) \cong E^{\lambda}$ (since $S^{\lambda}\cong\widetilde{M}^{\lambda}/Q^{\lambda}$). This proves Proposition 1.

Typos:

1. page 73, proof of Proposition 1 (§ 6.1): In the last identity, replace $\prod_{i=1}^m\left(1-x_jt\right)$ by $\prod_{i=1}^m\left(1-x_it\right)$.

2. page 91, proof of Theorem (§ 7.3): On line 4 of the proof, replace $M^{\left(\lambda\right)}$ by $M^\lambda$.

3. page 94, proof of Proposition 3 (§ 7.3): Three missing $)$ parentheses.

4. page 94, Exercise 11 (§ 7.3): Replace "$\mathbb{C}_{n}$" by "$\mathbb{C}^{n}$".

5. page 95, middle of the page (§ 7.4): "subspace generated by all $\left[T\right] - \operatorname{sgn}\left(q\right)\left[T\right]$" should be "subspace generated by all $\left[qT\right] - \operatorname{sgn}\left(q\right)\left[T\right]$".

6. page 101, proof of Proposition 4 (§ 7.4): The last sum ($\sum_{\ell=0}^k \left(-1\right)^\ell \dbinom{k}{\ell}$) should actually be $\sum_{\ell=0}^{k-m} \left(-1\right)^\ell \dbinom{k-m}{\ell}$. (Of course, this still equals $0$.)

7. page 103, Exercise 21 (§ 7.4): The sum $\sum_{\lambda, \nu}$ should actually be a sum $\sum_{\lambda, \mu}$.

8. page 111, Exercise 2 (§ 8.1): Of course, in (iii), the partition $\lambda$ should be assumed to be nonempty.

9. page 111, paragraph between Exercises 2 and 3 (§ 8.1): "a left action of the algebra $\operatorname{End}_R\left(E\right)$ on $E^\lambda$" should be "a left action of the multiplicative monoid $\operatorname{End}_R\left(E\right)$ on $E^\lambda$" (since this action is not linear in the acting element: i.e., two endomorphisms $U, V \in \operatorname{End}_R\left(E\right)$ will rarely satisfy $\left(U+V\right)\mathbf{e} = U\mathbf{e} + V\mathbf{e}$ for all $\mathbf{e} \in E^\lambda$). Likewise, the "$M_mR$" has to be understood as a multiplicative monoid, not as an algebra. Similarly, just below Exercise 3, "The algebra $M_mR$ also acts on the left on the $R$-algebra $R\left[Z\right]$" should be read as "The monoid $M_mR$ also acts by $R$-algebra endomorphisms on the left on the $R$-algebra $R\left[Z\right]$".

10. page 114, middle of the page (§ 8.2): "if and only if $\lambda _{i}+k=\lambda_{i}+k^{\prime}$ for all $i$" should be "if and only if $\lambda_{i}+k=\lambda_{i}^{\prime}+k^{\prime}$ for all $i$".

11. page 118, proof of Proposition 1 (§ 8.3): The $\lambda$ in "$E\left(S_\lambda\right)$" should be a superscript, not a subscript.

12. page 123, line -8 (§ 8.3): Replace "presentation" by "representation".

13. page 125, generators of $Q$ (§ 8.4): Both $w_p$s in the formula should be $w_q$s.

14. page 133, Exercise 1 (§ 9.1): The $i_{d+1}$ subscript should be $j_{d+1}$.

15. page 133, proof of Lemma 1 (§ 9.1): "Fix some $i_i, \ldots, i_d$" should be "Fix some $i_1, \ldots, i_d$", of course.

Answer 1

Here is a correct (I hope) proof of Lemma 4 in § 8.2 (inspired by § 5.3 in Desarmenien/Kung/Rota 1978):

Given an integer $1<k\leq m$ and a column $c$ of some tableau, we say that $c$ is $k$-defective if $c$ contains $k$ but not $k-1$. We also say that a column of a tableau is defective if it is $k$-defective for at least one $1<k\leq m$. It is easy to see that any non-defective column must have exactly the entries $1,2,\ldots,i$ (from top to bottom) for some $0\leq i\leq m$. Hence, the only tableau on $\lambda$ that has no defective columns is the tableau $U\left( \lambda\right) $. Hence, each tableau $T\neq U\left( \lambda\right) $ on $\lambda$ has at least one $k$-defective column for at least one $1<k\leq m$.

Let $w\in E^{\lambda}$ be a highest weight vector. We must show that $w$ is a scalar multiple of $e_{U\left( \lambda\right) }$.

Assume the contrary. Write $w$ in the form $w=\sum_{T}w_{T}e_{T}$, where $T$ varies over the tableaux on $\lambda$, and where the $w_{T}$ are coefficients in $\mathbb{C}$. This expression is unique by Theorem 1 in § 8.1.

We let $\mathfrak{T}$ be the set of all tableaux $T$ on $\lambda$ satisfying $w_{T}\neq0$. Thus, we can rewrite $w=\sum_{T}w_{T}e_{T}$ as $w=\sum _{T\in\mathfrak{T}}w_{T}e_{T}$.

Hence, at least one $T\in\mathfrak{T}$ is distinct from $U\left( \lambda\right) $ (since $w$ is not a scalar multiple of $e_{U\left( \lambda\right) }$). Thus, there exists at least one $1<k\leq m$ such that at least one tableau $T\in\mathfrak{T}$ has a $k$-defective column (since each tableau $T\neq U\left( \lambda\right) $ on $\lambda$ has at least one $k$-defective column for at least one $1<k\leq m$). Choose the smallest such $k$. Thus, \begin{equation} \text{no tableau }T\in\mathfrak{T}\text{ has a }p\text{-defective column for any }1<p<k. \label{eq.darij1.lem4.0} \tag{1} \end{equation}

Moreover, let $i$ be the maximum number of $k$-defective columns that a tableau $T\in\mathfrak{T}$ can have. Then, $i>0$ (since at least one tableau $T\in\mathfrak{T}$ has a $k$-defective column).

Now, let $t\in\mathbb{C}$ be arbitrary, and consider the upper triangular matrix $M:=I_{m}+tE_{k-1,k}\in B$, where $E_{p,q}$ is the elementary matrix with a $1$ in the $\left( p,q\right) $-th cell and zeroes everywhere else. The action of this matrix $M$ on $E$ transforms $e_{k}$ into $e_{k}+te_{k-1}$, while leaving all other basis vectors $e_{j}$ invariant.

But $M\in B$ and thus $Mw\in Bw=\mathbb{C}^{\ast}w$ (since $w$ is a highest weight vector). In other words, $Mw=\rho w$ for some $\rho\in\mathbb{C}^{\ast }$. Consider this $\rho$. We have $Mw=\rho w$. In view of $w=\sum _{T\in\mathfrak{T}}w_{T}e_{T}$, we can rewrite this as \begin{equation} \sum_{T\in\mathfrak{T}}w_{T}Me_{T}=\sum_{T\in\mathfrak{T}}\rho w_{T} e_{T}. \label{eq.darij1.lem4.1} \tag{2} \end{equation}

Let $T$ be a tableau on $\lambda$. Let $T\left[ k\rightarrow k-1\right] $ denote the result of replacing each entry $k$ of $T$ by a $k-1$, except for those $k$'s that already have a $k-1$ in their column. For example, if $k=3$ and $T= \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & 3 \\ 3 & 4 & 5 \end{bmatrix} $, then $T\left[ k\rightarrow k-1\right] = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 2 & 3 \\ 3 & 4 & 5 \end{bmatrix} $. It is easy to see that $T\left[ k\rightarrow k-1\right] $ is again a tableau. (Indeed, the entries of $T\left[ k\rightarrow k-1\right] $ clearly increase strictly in each column. To see that they increase weakly in each row, argue that the only thing that could go wrong would be a $k$ standing left of a $k-1$ in some row; but then the $k-1$ would have to come from a $k$ that got replaced by a $k-1$, but this would mean that the former $k$ has a $k-1$ in its column in $T$ but the latter $k$ does not, which easily violates the weakly-increasing property of the preceding row of $T$.)

Hence, if we recall how $e_{T}$ was defined (just before Lemma 1 in § 8.1), then we see that $Me_{T}$ is obtained from $e_{T}$ by replacing each $e_{k}$ (corresponding to a $k$ in $T$) by $e_{k}+te_{k-1}$. We can imagine this replacement happening gradually, column by column. When we replace a $k$ by $k-1$ in a given column of $T$, one of two things can happen: Either the $k$ that we replace already has a $k-1$ in its column (we call this a "non-defective $k$"), or it does not (we call this a "defective $k$"). In the former case, the replacement falls flat (i.e., changes nothing) (because $e_{k-1}\wedge\left( e_{k}+te_{k-1}\right) =e_{k-1}\wedge e_{k}$, so that the value of $e_{T}$ does not change when we replace that particular $e_{k}$ by $e_{k}+te_{k-1}$). In the latter case, $e_{k}$ becomes $e_{k}+te_{k-1}$, and upon expanding the tensor in $E^{\lambda}$ we obtain two addends, one of which is the one we had before our replacement while the other has an extra $t$ factor. At the end of our gradual replacement procedure, $e_{T}$ has become a sum of $2^{j\left( T\right) }$ addends of the form $e_{S}t^{s}$ (with $S$ being a filling, not always a tableau), where $j\left( T\right) $ is the total number of $k$-defective columns of $T$ (that is, the total number of defective $k$'s). Hence, expanding $Me_{T}$ as a polynomial in $t$, we find \begin{equation} Me_{T}=e_{T\left[ k\rightarrow k-1\right] }t^{j\left( T\right) }+\left( \text{some terms of the form }e_{S}t^{s}\text{ with }0<s<j\left( T\right) \right) +e_{T} \label{eq.darij1.lem4.2} \tag{3} \end{equation} (the highest power of $t$ comes from replacing every defective $k$ by $k-1$, whereas the constant term comes from replacing nothing). Note that the fillings $S$ on the right hand side are not always tableaux, but they all have fewer $k$'s than $T$ does, and therefore they can be written as linear combinations of $e_{Q}$'s for tableaux $Q$ that have fewer $k$'s than $T$ does (indeed, the first paragraph in the proof of Theorem 1 in § 8.1 explains how to rewrite $e_{S}$ for a filling $S$ as a linear combination of $e_{Q}$'s for tableaux $Q$, and it is easy to see that the latter $Q$'s have the same number of $k$'s as $S$).

Forget that we fixed $T$. Using the equation \eqref{eq.darij1.lem4.2}, which we just showed for each tableau $T$ on $\lambda$, we obtain \begin{align*} & \sum_{T\in\mathfrak{T}}w_{T}Me_{T}\\ & =\sum_{T\in\mathfrak{T}}w_{T}\left( e_{T\left[ k\rightarrow k-1\right] }t^{j\left( T\right) }+\left( \text{some terms of the form }e_{S} t^{s}\text{ with }0<s<j\left( T\right) \right) +e_{T}\right) \\ & =\sum_{T\in\mathfrak{T}\text{ with }j\left( T\right) =i}w_{T}e_{T\left[ k\rightarrow k-1\right] }t^{i}+\left( \text{some terms of the form } e_{S}t^{s}\text{ with }0<s<i\right) \\ & \ \ \ \ \ \ \ \ \ \ +\sum_{T\in\mathfrak{T}}w_{T}e_{T} \end{align*} (since the maximum $j\left( T\right) $ for a $T\in\mathfrak{T}$ is $i$ (by the definition of $i$)). In view of \eqref{eq.darij1.lem4.1}, we can rewrite this as \begin{align} \sum_{T\in\mathfrak{T}}\rho w_{T}e_{T} & =\underbrace{\sum_{T\in \mathfrak{T}\text{ with }j\left( T\right) =i}w_{T}e_{T\left[ k\rightarrow k-1\right] }t^{i}+\left( \text{some terms of the form }e_{S}t^{s}\text{ with }0<s<i\right) }_{\text{we call these terms the "error terms"}}\nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum_{T\in\mathfrak{T}}w_{T}e_{T} . \label{eq.darij1.lem4.4} \tag{4} \end{align}

Recall that the $e_{T}$'s for all tableaux $T$ on $\lambda$ are linearly independent (by Theorem 1 in § 8.1). Hence, comparing coefficients in front of $e_{T}$ for a strategically chosen $T$ (namely, some tableau $T\in \mathfrak{T}$ with the largest number of $k$'s in it), we conclude from \eqref{eq.darij1.lem4.4} that $\rho=1$ (because this $T$ only appears in the sum on the left hand side and in the very last sum on the right hand side, but not among the "error terms", because the $S$'s in the error terms have fewer $k$'s than $T$ does). Hence, \eqref{eq.darij1.lem4.4} rewrites as \begin{align} \sum_{T\in\mathfrak{T}}w_{T}e_{T} & =\sum_{T\in\mathfrak{T}\text{ with }j\left( T\right) =i}w_{T}e_{T\left[ k\rightarrow k-1\right] } t^{i}+\left( \text{some terms of the form }e_{S}t^{s}\text{ with }0<s<i\right) \nonumber\\ & \ \ \ \ \ \ \ \ \ \ +\sum_{T\in\mathfrak{T}}w_{T}e_{T} . \label{eq.darij1.lem4.5} \tag{5} \end{align}

Now, forget that we fixed $t$. We thus have proved the equality \eqref{eq.darij1.lem4.5} for each $t\in\mathbb{C}$. But \eqref{eq.darij1.lem4.5} is a polynomial identity in $t$ (with coefficients in $E^{\lambda}$), and thus must hold as a formal equality between polynomials if it holds for each $t\in\mathbb{C}$ (since $\mathbb{C}$ is an infinite field). Hence, we can compare coefficients in \eqref{eq.darij1.lem4.5}. In particular, comparing coefficients in front of $t^{i}$, we conclude that \begin{equation} 0=\sum_{T\in\mathfrak{T}\text{ with }j\left( T\right) =i}w_{T}e_{T\left[ k\rightarrow k-1\right] } \label{eq.darij1.lem4.6} \tag{6} \end{equation} (since $i>0$). Note that there exists at least one $T\in\mathfrak{T}$ with $j\left( T\right) =i$ (by the definition of $i$), so that the sum on the right hand side of \eqref{eq.darij1.lem4.6} is nonempty.

But the $e_{S}$ for different tableaux $S$ on $\lambda$ are linearly independent (by Theorem 1 in § 8.1). Thus, the equality \eqref{eq.darij1.lem4.6} is only possible if there exist two different $T\in\mathfrak{T}$ with $j\left( T\right) =i$ that have the same $T\left[ k\rightarrow k-1\right] $ (since the sum on the right hand side of \eqref{eq.darij1.lem4.6} is nonempty, and all coefficients $w_{T}$ are nonzero by the definition of $\mathfrak{T}$). So this must be the case. In other words, there exist two distinct tableaux $T_{1}$ and $T_{2}$ in $\mathfrak{T}$ with $j\left( T_{1}\right) =j\left( T_{2}\right) =i$ and $T_{1}\left[ k\rightarrow k-1\right] =T_{2}\left[ k\rightarrow k-1\right] $.

But I claim that such $T_{1}$ and $T_{2}$ in $\mathfrak{T}$ cannot exist. In other words, I claim that any $T\in\mathfrak{T}$ with $j\left( T\right) =i$ can be uniquely reconstructed from $T\left[ k\rightarrow k-1\right] $. How? Let me show:

Recall that $T\left[ k\rightarrow k-1\right] $ is constructed from $T$ by replacing all $i$ "defective" $k$'s by $k-1$'s. Thus, we could reconstruct $T$ from $T\left[ k\rightarrow k-1\right] $ by replacing them back, if we knew which of the $k-1$'s in $T\left[ k\rightarrow k-1\right] $ are coming from $k$'s in $T$. Let us try to find them.

Any entry $k-1$ in $T$ must have the entries $1,2,\ldots,k-2$ above it (since otherwise, its column would be $p$-defective for some $1<p<k$, but this would contradict \eqref{eq.darij1.lem4.0}), and thus must lie in the $\left( k-1\right) $-st row. In other words, any entry of $T$ that does not lie in the $\left( k-1\right) $-st row cannot be a $k-1$.

Hence, any entry $k-1$ in $T\left[ k\rightarrow k-1\right] $ that does not lie in the $\left( k-1\right) $-st row cannot come from a $k-1$ in $T$, but must have come from a $k$ in $T$ that was replaced by a $k-1$. Thus, in order to reconstruct $T$, we must replace any such $k-1$ back by $k$.

It remains to figure out what to do with the $k-1$'s in $T\left[ k\rightarrow k-1\right] $ that do lie in the $\left( k-1\right) $-st row: Which of them should we replace back by $k$'s and which should we leave as they are? Fortunately, we know how many of them we should replace: We have obtained $T\left[ k\rightarrow k-1\right] $ from $T$ by replacing $j\left( T\right) =i$ many $k$'s by $k-1$'s, and thus we must (in turn) replace $i$ many $k-1$'s by $k$'s to reconstruct $T$. Let $s$ be the number of other $k-1$'s that we have already replaced back by $k$'s (these are the ones that do not lie in the $\left( k-1\right) $-st row). Then, we still need to replace $i-s$ many $k-1$'s in the $\left( k-1\right) $-st row back by $k$'s. Since our goal is to obtain a tableau (because $T$ is a tableau), these $i-s$ many $k-1$'s must be the rightmost $k-1$'s in that row. Thus, we replace the rightmost $i-s$ many $k-1$'s in the $\left( k-1\right) $-st row by $k$'s. At that point, we have reconstructed $T$.

As explained, this shows that two distinct tableaux $T_{1}$ and $T_{2}$ in $\mathfrak{T}$ with $j\left( T_{1}\right) =j\left( T_{2}\right) =i$ and $T_{1}\left[ k\rightarrow k-1\right] =T_{2}\left[ k\rightarrow k-1\right] $ cannot exist. This contradicts their existence that was proved before, and so the proof is complete.