The joint spectrum of the Jucys-Murphy elements is the set of tableau content vectors

For a standard tableau $T$, let ${\lambda }^{(k)}$ be the shape of its entries $1,\dots ,k$. The Young line $L_T$ lies in the corresponding irreducible summand at each level of the restriction chain by [F2] and the given diagonal-algebra construction. For $k\ge 2$, $T_k$ and $T_{k-1}$ therefore act on this line by $z_{{\lambda }^{(k)}}$ and $z_{{\lambda }^{(k-1)}}$. Since $X_k=T_k-T_{k-1}$, [F1] gives its eigenvalue as their difference, the content of the one node added at step $k$. For $k=1$, $X_1=0$ and entry $1$ occupies $(1,1)$, of content zero. Thus $X_k{v}_T=c_T(k)v_T$ for all $k$.

Every tableau content vector satisfies (1) and (2): entry $1$ lies in $(1,1)$, and any later node has a left or upper neighbour of smaller entry and content respectively one less or one greater. Nodes of a fixed content lie on a northwest-to-southeast diagonal, and their entries strictly increase along it: the right neighbour of an earlier diagonal node lies before the next diagonal node. If entries $p<q$ have equal content, their nodes are $(r,c)$ and $(r+d,c+d)$ for $d\ge 1$. The nodes $(r,c+1)$ and $(r+1,c)$ exist because the later node does, and their entries lie strictly between $p$ and $q$ by row/column increase along paths inside the diagram. Their contents are respectively $c-r+1$ and $c-r-1$, proving (3).

We prove the precise criterion needed to construct a tableau. Let a nonempty standard tableau have content vector $\beta$ and shape $\lambda$. An integer $t$ is an addable content exactly when (A) $t-1$ or $t+1$ occurs in $\beta$, and (B) after every occurrence of $t$, both neighbours $t-1,t+1$ occur later in $\beta$. First suppose the diagonal of content $t$ is absent. Then $t\mathrm{\ne }0$. If $t>0$, absence means ${\lambda }_1\le t$; content $t+1$ is also absent, and content $t-1$ occurs exactly when ${\lambda }_1\ge t$. Thus (A) says ${\lambda }_1=t$, exactly when $(1,t+1)$ is addable. If $t<0$, transpose the diagram: absence means the first column has height at most $-t$, and (A) says its height is exactly $-t$, exactly when the new bottom node of content $t$ is addable. In these cases (B) is vacuous.

Suppose instead that the last node of content $t$ is $x=(r,c)$, with $c-r=t$. Any addable node of that content must be $(r+1,c+1)$: a diagram is closed under moving northwest, so all earlier nodes on the same diagonal already exist and a later one would require its immediate predecessor. This next node is addable exactly when both $u=(r,c+1)$ and $v=(r+1,c)$ exist. If they exist, each entry is larger than the entry of $x$ and their contents are $t+1,t-1$, so (A) and (B) hold, since $x$ has the largest entry among the content-$t$ nodes. Conversely, if $u$ is absent, every content-$(t+1)$ node $(a,a+t+1)$ has $a<r$: a node with $a>r$ would force a content-$t$ node in its own row beyond $x$, and $a=r$ would be $u$. Its column is then at most $c$, so it is northwest of $x$ and has smaller entry. If $v$ is absent, every content-$(t-1)$ node $(a,a+t-1)$ has $a\le r$: $a\ge r+2$ would force the content-$t$ node $(a-1,a+t-1)$ beyond $x$, while $a=r+1$ would be $v$. Such a node is again strictly northwest of $x$ and has smaller entry. Thus in either absence case (B) fails at the entry of $x$. This proves the criterion completely.

Given $\alpha$ satisfying (1)-(3), start with entry $1$ at $(1,1)$. Suppose its first $q-1$ entries have been placed in a standard tableau. Condition (2) at $q$ is (A) of the criterion, and condition (3) applied to every earlier occurrence of $a_q$ is exactly (B). Steps 2.1 and 3.1 therefore give an addable node of content $a_q$. It is unique by [F3]. Put entry $q$ in that node; the shape remains a Young diagram and standardness holds because every previous entry is smaller. Induction constructs a tableau with content vector $\alpha$. Any tableau with that vector has the same successive shapes and entries, by uniqueness of the addable node at each step, so it is the same tableau. Combined with step 1.2, this proves the asserted bijection.

The Young lines give a basis in each $S_{\mathbb{C}}^{\lambda }$ by [F4] and the given diagonal-algebra theorem; their union is a basis of the specified multiplicity-free sum. By step 1.1 their joint weights are precisely the tableau content vectors. Step 4.1 proves these vectors are distinct: across all shapes, each vector specifies exactly one tableau. In a basis of joint eigenvectors, the eigenspace for a fixed vector is the span of exactly those basis vectors with that weight, because comparing each coordinate coefficient in $X_{k}v=a_{k}v$ forces any nonzero coefficient to have that weight for every $k$. Hence each such eigenspace is one-dimensional, and the eigenvalue vectors are exactly the integer vectors satisfying (1)-(3). This proves (a) and (b).