Schur-Weyl for two tensor factors

Put $p_{+}:=\frac{1+\tau }{2}$ and $p_{-}:=\frac{1-\tau }{2}$ in $\mathbb{C}[S_2]$. Using ${\tau }^2=\mathrm{i}\mathrm{d}$ one computes $p_{+}+p_{-}=\mathrm{i}\mathrm{d}$, $p_{+}^2=p_{+}$, $p_{-}^2=p_{-}$ and $p_{+}{p}_{-}=p_{-}{p}_{+}=0$; hence $E=p_{+}E\oplus p_{-}E$, and $p_{+}E$ is the fixed space ${\mathrm{Sym}}^{2}V$ while $p_{-}E$ is the anti-fixed space ${\mathrm{\Lambda }}^{2}V$.

The two rank-$2$ Specht modules are the expected one-dimensional modules. For $\lambda =(2)$ every $(2)$-tableau has the same tabloid $\{t\}=\{1,2\}$ and trivial column stabilizer, so $e_s=\{t\}$ for every such tableau and $S^{(2)}=\mathbb{C}\{t\}$ with $\tau \cdot \{t\}=\{\tau \cdot t\}=\{t\}$: the action is trivial. For $\lambda =(1,1)$ the two tabloids $\{u\}$ and $\tau \cdot \{u\}$ are distinct basis vectors of $M^{(1,1)}$; with $C_u=\{\mathrm{i}\mathrm{d},\tau \}$ one has ${\kappa }_u=\mathrm{i}\mathrm{d}-\tau$, so $e_u=\{u\}-\tau \cdot \{u\}\mathrm{\ne }0$, and $\tau \cdot e_u=\tau \cdot \{u\}-\{u\}=-e_u$; by [F3] any one polytabloid generates $S^{(1,1)}$, so $S^{(1,1)}=\mathbb{C}{e}_u$ is the sign representation, of dimension one. In particular $S^{(2)}$ is trivial and $S^{(1,1)}$ is the sign representation, as claimed in the Statement.

The two spaces are spanned by explicit tensors: set $s_{ii}:=e_i\otimes e_i$ and $s_{ij}:=e_i\otimes e_j+e_j\otimes e_i$ for $i<j$, and set $a_{ij}:=e_i\otimes e_j-e_j\otimes e_i$ for $i<j$. If $x={\sum }_{i,j}{c}_{ij}{e}_i\otimes e_j$ is fixed by $\tau$, then $c_{ij}=c_{ji}$, so $x={\sum }_i{c}_{ii}{s}_{ii}+{\sum }_{i<j}{c}_{ij}{s}_{ij}$. If $x$ is anti-fixed, then $c_{ij}=-c_{ji}$ and $2c_{ii}=0$; over $\mathbb{C}$ this gives $c_{ii}=0$, so $x={\sum }_{i<j}{c}_{ij}{a}_{ij}$. Conversely each $s_{ij}$ is fixed and each $a_{ij}$ is anti-fixed. Their respective coefficients on the tensor basis [F2] show that both displayed families are linearly independent: for the symmetric family use the coefficient of $e_i\otimes e_i$ for $s_{ii}$ and of $e_i\otimes e_j$ for $s_{ij}$ with $i<j$; for the alternating family use the coefficient of $e_i\otimes e_j$ for each $i<j$. Hence $\dim {\mathrm{Sym}}^{2}V=\left(\genfrac{}{}{0px}{}{d+1}{2}\right)=\frac{d(d+1)}{2}$ and $\dim {\mathrm{\Lambda }}^{2}V=\left(\genfrac{}{}{0px}{}{d}{2}\right)=\frac{d(d-1)}{2}$, and step 1.1 gives the direct sum $E={\mathrm{Sym}}^{2}V\oplus {\mathrm{\Lambda }}^{2}V$.

Both summands are $S_2$-submodules: $S_2=\{\mathrm{i}\mathrm{d},\tau \}$, and each of ${\mathrm{Sym}}^{2}V$ and ${\mathrm{\Lambda }}^{2}V$ is stable under $\tau$ by its definition, hence under every element of $S_2$.

By [F4] and the list of partitions of $2$ there is an $(S_2\times \mathrm{GL}(V))$-isomorphism $E\cong S^{(2)}\otimes M_{(2)}\oplus S^{(1,1)}\otimes M_{(1,1)}$ when $d\ge 2$, and $E\cong S^{(2)}\otimes M_{(2)}$ when $d\le 1$ (for $d=0$ the sum is empty and $E=0$). By step 2.1 the transposition acts as $+1$ on $S^{(2)}$ and as $-1$ on $S^{(1,1)}$; since $S_2$ acts trivially on the multiplicity spaces, $\tau$ acts as $+1$ on $S^{(2)}\otimes M_{(2)}$ and as $-1$ on $S^{(1,1)}\otimes M_{(1,1)}$.

Consequently $S^{(2)}\otimes M_{(2)}\subseteq {\mathrm{Sym}}^{2}V$ and $S^{(1,1)}\otimes M_{(1,1)}\subseteq {\mathrm{\Lambda }}^{2}V$; since by step 1.1 the whole of $E$ is the direct sum of its $\tau$-fixed and $\tau$-anti-fixed parts, these inclusions are equalities: the $\tau$-fixed part of $E$ is exactly the $(2)$-summand and the $\tau$-anti-fixed part is exactly the $(1,1)$-summand. In particular, when $d\le 1$ the $(1,1)$-summand is absent, so the whole of $E$ is $\tau$-fixed and ${\mathrm{\Lambda }}^{2}V=0$; when $d\ge 2$ both summands are nonzero.

Reading off dimensions in the equality ${\mathrm{Sym}}^{2}V=S^{(2)}\otimes M_{(2)}$ of step 4.1 and using ${\dim }_{\mathbb{C}}{S}^{(2)}=1$ from step 2.1 gives ${\dim }_{\mathbb{C}}{M}_{(2)}=\frac{d(d+1)}{2}$ for $d\ge 1$ (and $M_{(2)}=0$ for $d=0$, when $E=0$); similarly ${\dim }_{\mathbb{C}}{M}_{(1,1)}=\frac{d(d-1)}{2}$ for $d\ge 2$, while $M_{(1,1)}=0$ for $d\le 1$; this is exactly the length cutoff $\mathrm{\ell }((1,1))=2>d$ for $d\le 1$. Since $\frac{d(d-1)}{2}>0$ precisely when $d\ge 2$, the alternating square vanishes if and only if $d<2$, and the two extreme cases are $d=0$ (both spaces zero, empty Schur-Weyl sum) and $d=1$ ($E={\mathrm{Sym}}^{2}V$ one-dimensional, ${\mathrm{\Lambda }}^{2}V=0$). The two idempotents of step 1.1 use that $2$ is invertible in $\mathbb{C}$, so the calculation is specific to characteristic zero; all arguments use the fixed basis and the explicit two-element group, and no choice principle is invoked. This proves the Statement.