15.7  Symmetric Powers

Proof. Word for word the proof of Theorem 15.6.1, with “symmetric $k$-linear” in place of “$k$-linear” throughout: (S1) for $P$ applied to the symmetric map ${\pi }^{\prime }$ gives $\phi$, (S1) for $P^{\prime }$ applied to $\pi$ gives $\psi$, the composite $\psi \phi$ is a linear self-map of $P$ with $\psi \phi \pi =\pi$, and the uniqueness clause of (S1) applied to $\pi$ forces $\psi \phi ={\mathrm{id}}_P$. Symmetrically $\phi \psi ={\mathrm{id}}_{P^{\prime }}$. This proves the theorem.

Proof. $C_k$ is a subspace, being a span. The map $\pi$ is $k$-linear, being the $k$-linear ${\mu }_k$ of Theorem 15.6.2 followed by the linear map $q$. It is symmetric: for $\sigma \in S_k$, the difference $$u_1\otimes \cdots \otimes u_k-u_{\sigma (1)}\otimes \cdots \otimes u_{\sigma (k)}$$ lies in $C_k$ by definition, so the two tensors have the same image under $q$: $\pi (u_{\sigma (1)},\dots ,u_{\sigma (k)})=\pi (u_1,\dots ,u_k)$.

Now let $f:V^k\to W$ be symmetric $k$-linear. By (T1) there is a unique linear $\tilde{f}:V^{\otimes k}\to W$ with $\tilde{f}(u_1\otimes \cdots \otimes u_k)=f(u_1,\dots ,u_k)$. On a generator of $C_k$, $$\tilde{f}(u_1\otimes \cdots \otimes u_k-u_{\sigma (1)}\otimes \cdots \otimes u_{\sigma (k)})=f(u_1,\dots ,u_k)-f(u_{\sigma (1)},\dots ,u_{\sigma (k)}),$$ which is $0$ because $f$ is symmetric. A linear map vanishing on a spanning set of $C_k$ vanishes on $C_k$, so $C_k\subseteq \ker \tilde{f}$. By Theorem 4.4.10 (a) there is exactly one linear $\overline{f}:V^{\otimes k}/C_k\to W$ with $\overline{f}\circ q=\tilde{f}$, and then $$\overline{f}\circ \pi =\overline{f}\circ q\circ {\mu }_k=\tilde{f}\circ {\mu }_k=f.$$

For uniqueness, suppose $h$ is linear with $h\circ \pi =f$. Since $q$ is surjective and $V^{\otimes k}$ is spanned by its simple tensors (Theorem 15.6.2), the elements $\pi (u_1,\dots ,u_k)$ span $V^{\otimes k}/C_k$. So $h$ and $\overline{f}$ agree on a spanning set, hence everywhere. This proves the theorem.

Proof. Spanning. By Theorem 15.6.3 the tensors $v_{j_1}\otimes \cdots \otimes v_{j_k}$, over all tuples $J$, form a basis of $V^{\otimes k}$, and $q$ is surjective, so their images $v_{j_1}\cdots v_{j_k}$ span ${\mathrm{Sym}}^{k}V$. Given any $J$, let $\sigma \in S_k$ sort it, so that $(j_{\sigma (1)},\dots ,j_{\sigma (k)})$ is nondecreasing. Since $\pi$ is symmetric, $v_{j_1}\cdots v_{j_k}=v_{j_{\sigma (1)}}\cdots v_{j_{\sigma (k)}}$. So the nondecreasing tuples already give a spanning set.

Independence. Let $({\phi }_1,\dots ,{\phi }_n)$ be the dual basis of $\mathcal{B}$ (Theorem 5.1.3). For a nondecreasing tuple $I=(i_1,\dots ,i_k)$, let $$m_j=|\{t:i_t=j\}|,j=1,\dots ,n,$$ be its multiplicity vector, so $m_1+\cdots +m_n=k$, and let $\mathrm{\Phi }(I)$ be the set of functions $\tau :\{1,\dots ,k\}\to \{1,\dots ,n\}$ taking the value $j$ exactly $m_j$ times, for each $j$. Define $$g_I(u_1,\dots ,u_k)=\sum _{\tau \in \mathrm{\Phi }(I)}{\phi }_{\tau (1)}(u_1){\phi }_{\tau (2)}(u_2)\cdots {\phi }_{\tau (k)}(u_k).$$ Each summand is $k$-linear, being a product of one functional per slot, so $g_I$ is $k$-linear.

It is symmetric. Let $\rho \in S_k$. Substituting and then reindexing the product by $s=\rho (t)$, $$g_I(u_{\rho (1)},\dots ,u_{\rho (k)})=\sum _{\tau \in \mathrm{\Phi }(I)}\prod _{s=1}^k{\phi }_{(\tau {\rho }^{-1})(s)}(u_s).$$ The map $\tau \mapsto \tau {\rho }^{-1}$ is a bijection of $\mathrm{\Phi }(I)$ onto itself, since composing with a permutation of the domain does not change how often a value is taken. So the sum is unchanged, and $g_I$ is symmetric.

By (S1), $g_I$ induces a linear ${\overline{g}}_I:{\mathrm{Sym}}^{k}V\to F$. Evaluate it on a basis candidate: for any tuple $J$, $${\overline{g}}_I(v_{j_1}\cdots v_{j_k})=\sum _{\tau \in \mathrm{\Phi }(I)}\prod _{t=1}^k{\delta }_{\tau (t)j_t}.$$ A summand is $1$ exactly when $\tau (t)=j_t$ for every $t$, and $0$ otherwise; so the whole sum is $1$ if the function $t\mapsto j_t$ belongs to $\mathrm{\Phi }(I)$, and $0$ if it does not. For nondecreasing $I$ and $J$, that function lies in $\mathrm{\Phi }(I)$ exactly when $J$ has the same multiplicity vector as $I$, which for nondecreasing tuples means $J=I$. Hence ${\overline{g}}_I(v_{j_1}\cdots v_{j_k})={\delta }_{IJ}$.

Now suppose $\sum _I{a}_I{v}_{i_1}\cdots v_{i_k}=0$, the sum over nondecreasing tuples. Applying ${\overline{g}}_J$ leaves $a_J=0$. Since $J$ was arbitrary, every coefficient is $0$, and the list is independent. So it is a basis.

The count. Send a nondecreasing tuple $(i_1,\dots ,i_k)$ to $$(i_1,i_2+1,i_3+2,\dots ,i_k+k-1).$$ Consecutive entries satisfy $i_t+t-1<i_{t+1}+t$ because $i_t\le i_{t+1}$, so the image is strictly increasing, and its entries lie between $1$ and $n+k-1$. Conversely, subtracting $0,1,\dots ,k-1$ from a strictly increasing tuple in $\{1,\dots ,n+k-1\}$ returns a nondecreasing tuple in $\{1,\dots ,n\}$: the differences $j_{t+1}-j_t\ge 1$ become $i_{t+1}-i_t\ge 0$, and $1\le j_1$, $j_k\le n+k-1$ give $1\le i_1$ and $i_k\le n$. The two assignments are mutually inverse, so the nondecreasing $k$-tuples from $\{1,\dots ,n\}$ are in bijection with the $k$-element subsets of $\{1,\dots ,n+k-1\}$, and there are $(\genfrac{}{}{0ex}{}{n+k-1}{k})$ of those. This proves the theorem.

Proof. Concatenation carries the relations into the relations. Let $x\in C_k$ and $y\in V^{\otimes l}$. For a generator $x=u_1\otimes \cdots \otimes u_k-u_{\sigma (1)}\otimes \cdots \otimes u_{\sigma (k)}$ and a simple $y=w_1\otimes \cdots \otimes w_l$, the product $x\cdot y$ of Proposition 15.6.4 is again a generator of $C_{k+l}$, for the permutation acting as $\sigma$ on the first $k$ slots and fixing the last $l$. Since generators span $C_k$, since $V^{\otimes l}$ is spanned by its simple tensors, and since $\cdot$ is bilinear, $C_k\cdot V^{\otimes l}\subseteq C_{k+l}$. Swapping the roles of the two blocks gives $V^{\otimes k}\cdot C_l\subseteq C_{k+l}$.

Descending, one argument at a time. Let $q_m:V^{\otimes m}\to {\mathrm{Sym}}^{m}V$ denote the quotient maps. Fix $y\in V^{\otimes l}$. The linear map $x\mapsto q_{k+l}(x\cdot y)$ kills $C_k$ by the first inclusion, so Theorem 4.4.10 gives a unique linear ${\beta }_y:{\mathrm{Sym}}^{k}V\to {\mathrm{Sym}}^{k+l}V$ with ${\beta }_y(q_{k}x)=q_{k+l}(x\cdot y)$. The assignment $y\mapsto {\beta }_y$ is linear, since ${\beta }_{ay+y^{\prime }}$ and $a{\beta }_y+{\beta }_{y^{\prime }}$ agree on the spanning set $q_k(V^{\otimes k})={\mathrm{Sym}}^{k}V$; and it kills $C_l$, since $y\in C_l$ gives $x\cdot y\in C_{k+l}$ and hence ${\beta }_y=0$. By Theorem 4.4.10 again it descends to a linear map ${\mathrm{Sym}}^{l}V\to \mathcal{L}({\mathrm{Sym}}^{k}V,{\mathrm{Sym}}^{k+l}V)$, and evaluating gives the required bilinear map. It is unique because the products of vectors span each factor. When $k=0$ or $l=0$ the product is scalar multiplication, as in Section 06. This proves the proposition.

Proof. Bilinearity, associativity, and the identity are proved exactly as in Theorem 15.6.5: all three are identities between multilinear expressions, so it suffices to check them on products of vectors, which span each ${\mathrm{Sym}}^{k}V$, and there all three read “concatenate the strings”.

Commutativity is the new point. Both $(u_1\cdots u_k)(w_1\cdots w_l)$ and $(w_1\cdots w_l)(u_1\cdots u_k)$ equal the symmetric product of the same $k+l$ vectors in two different orders, and $\pi$ is symmetric, so they are equal. Two bilinear maps agreeing on spanning sets of both arguments agree everywhere, so $xy=yx$ for all $x,y$. This proves the proposition.

Proof. Write $|m|=m_1+\cdots +m_n$ for $m\in {\mathbb{N}}^n$. Given $m$ with $|m|=k$, let $I(m)$ be the nondecreasing tuple taking the value $j$ exactly $m_j$ times, and write $v^m$ for the corresponding symmetric product $v_{i_1}\cdots v_{i_k}$. The assignment $m\mapsto I(m)$ is a bijection onto the nondecreasing $k$-tuples, since a nondecreasing tuple is recovered from its multiplicity vector and conversely. So by Theorem 15.7.3 the family $(v^m{)}_{|m|=k}$ is a basis of ${\mathrm{Sym}}^{k}V$, and hence $(v^m{)}_{m\in {\mathbb{N}}^n}$ is a basis of $\mathrm{Sym}(V)$.

Define $\mathrm{\Psi }$ to be the unique linear map with $\mathrm{\Psi }(v^m)=x^m$, which exists by Theorem 2.1.6. It sends a basis of $\mathrm{Sym}(V)$ bijectively onto the basis of monomials, so it is a linear isomorphism, and it sends the degree-$k$ basis onto the degree-$k$ monomials, so $\mathrm{\Psi }({\mathrm{Sym}}^{k}V)=F[x_1,\dots ,x_n{]}_k$.

Multiplicativity: the maps $(y,z)\mapsto \mathrm{\Psi }(yz)$ and $(y,z)\mapsto \mathrm{\Psi }(y)\mathrm{\Psi }(z)$ are bilinear, so it is enough to compare them on the basis. By Proposition 15.7.4, $v^m{v}^{m^{\prime }}$ is the symmetric product of the $|m|+|m^{\prime }|$ vectors of the two strings, which after sorting is $v^{m+m^{\prime }}$; and $x^m{x}^{m^{\prime }}=x^{m+m^{\prime }}$ by Definition 15.7.4. Also $\mathrm{\Psi }$ sends the empty product $1$ to the monomial with all exponents $0$, which is $1$. Hence $\mathrm{\Psi }$ is an algebra isomorphism. Finally $v_i$ is the basis element $v^m$ whose $m$ has a single $1$ in position $i$, so $\mathrm{\Psi }(v_i)=x_i$.

The dimension formula follows from Theorem 15.7.3, since $\mathrm{\Psi }$ restricts to an isomorphism in each degree. This proves the theorem.

Proof. The map $(u_1,\dots ,u_k)\mapsto u_{\sigma (1)}\otimes \cdots \otimes u_{\sigma (k)}$ is $k$-linear, since in slot $t$ it is ${\mu }_k$ in slot ${\sigma }^{-1}(t)$. So (T1) gives a unique linear $P_{\sigma }$ with the stated values, and uniqueness holds because those tensors span $V^{\otimes k}$. For the composition rule, put $w_t=u_{\tau (t)}$; then $$P_{\sigma }{P}_{\tau }(u_1\otimes \cdots \otimes u_k)=P_{\sigma }(w_1\otimes \cdots \otimes w_k)=w_{\sigma (1)}\otimes \cdots \otimes w_{\sigma (k)},$$ and $w_{\sigma (t)}=u_{\tau (\sigma (t))}$, which is the $t$-th argument of $P_{\tau \sigma }$. Both sides are linear and agree on simple tensors, hence are equal. This proves the lemma.

Proof.