Vanishing on products

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

By the diagonal criterion (Proposition 3.1 of Symmetric powers and the Foulkes–Howe map), Theorem A follows from a vanishing statement for symmetric multilinear forms. This lesson proves it by induction on the number of slots, following OpenAI [OpenAI-FH, Section 4]. Two sequences of directional derivatives do the work, and each is undone by the injectivity lemma of A differential injectivity lemma: the first frees one slot from a common factor, and the second, after the induction has made the other slots independent, removes the common power that remains. The second sequence raises the degree of one variable \(r^2\) times, which is where the quadratic bound comes from. Section 3 derives the equivariant embedding of Corollary B.

Throughout, \(V\) is a finite-dimensional complex vector space, products are taken in \(\operatorname{Sym}V\), and a product of \(k\) vectors is \(u_1\cdots u_k\) with \(u_i\in V\).

1. The vanishing theorem

Proposition 1.1 (OpenAI). Let \(a,b\ge1\) with \(b\ge a(a-1)\). If a symmetric \(a\)-linear form \(T\) on \(\operatorname{Sym}^bV\) satisfies \(T(P,\dots,P)=0\) for every product \(P\) of \(b\) vectors, then \(T=0\).

Proof. If \(V=0\), then \(\operatorname{Sym}^bV=0\). Let \(\dim V=n\ge1\), and fix a basis, so that polynomials in vector variables are defined as in A differential injectivity lemma. We prove the statement by induction on \(a\), for all admissible \(b\) and all \(V\) at once. For \(a=1\), \(T\) is a linear form vanishing on all products of \(b\) vectors, which span \(\operatorname{Sym}^bV\).

Let \(a\ge2\) and assume the statement for \(a-1\). Put \(r=a-1\) and \(m=b-r\). Then \(b\ge(r+1)r\) gives \(m\ge r^2\ge1\).

Step 1: a split binary form. Let \(Q=y_1\cdots y_m\) be a product of \(m\) vectors, and \(x,t\in V\). For \(c\in\mathbb C\), \(c\ne0\), choose \(\alpha\) with \(\alpha^r=-c\) and let \(\zeta=e^{2\pi i/r}\); then \[ X^r+cY^r=\prod_{j=0}^{r-1}(X-\alpha\zeta^jY) \] as polynomials in \(X,Y\), because \(\prod_j(X-\zeta^jZ)=X^r-Z^r\) (the \(\zeta^j\) are the \(r\)-th roots of unity), applied with \(Z=\alpha Y\). Substituting \(X=x\), \(Y=t\) in \(\operatorname{Sym}V\) shows that \((x^r+ct^r)Q\) is a product of \(r+m=b\) vectors; for \(c=0\) it is \(x^rQ\). So the polynomial in \(c\) \[ T\bigl((x^r+ct^r)Q,\dots,(x^r+ct^r)Q\bigr) \] vanishes for all \(c\), and so do its coefficients. By symmetry and multilinearity the coefficient of \(c\) is \(a\,T(x^rQ,\dots,x^rQ,t^rQ)\), with \(r\) slots \(x^rQ\). Hence \[ T(x^rQ,\dots,x^rQ,t^rQ)=0 .\tag{1} \]

Step 2: freeing the last slot. Fix \(x\in V\), let \(t,y_1,\dots,y_m\) be vector variables, \(Q=y_1\cdots y_m\), and \[ H=T(x^rQ,\dots,x^rQ,t^b), \] with \(r\) slots \(x^rQ\). It is homogeneous of degree \(b\) in \(t\) and of degree \(r\) in each \(y_i\). Only the last slot involves \(t\), so by Section 1 of A differential injectivity lemma, for \(0\le k\le m\), \[ \begin{aligned} &D_{y_k,t}\cdots D_{y_1,t}H=\frac{b!}{(b-k)!}\\ &\quad\times T(x^rQ,\dots,x^rQ,t^{b-k}y_1\cdots y_k). \end{aligned} \] For \(k=m\) the last slot is \(t^rQ\), and (1) shows that the right side vanishes at every point, hence is the zero polynomial. Immediately before the \(i\)-th operator \(D_{y_i,t}\), the polynomial has degree \(b-i+1\) in \(t\) and degree \(r\) in \(y_i\), since the earlier operators raised only the degrees of \(y_1,\dots,y_{i-1}\); and \(b-i+1-r=m-i+1\ge1\). By Lemma 3.1 of A differential injectivity lemma each operator is injective there, so \(H=0\). Hence, for all \(x,t\in V\) and all products \(Q\) of \(m\) vectors, \[ T(x^rQ,\dots,x^rQ,t^b)=0 .\tag{2} \]

Step 3: the induction. For fixed \(x,t\in V\), \[ S(Q_1,\dots,Q_r)=T(x^rQ_1,\dots,x^rQ_r,t^b) \] is a symmetric \(r\)-linear form on \(\operatorname{Sym}^mV\), and by (2) it vanishes on the diagonal at every product of \(m\) vectors. As \(m\ge1\) and \(m\ge r^2\ge r(r-1)\), the induction hypothesis gives \(S=0\): for all \(Q_1,\dots,Q_r\in\operatorname{Sym}^mV\), \[ T(x^rQ_1,\dots,x^rQ_r,t^b)=0 .\tag{3} \]

Step 4: removing the common power. Fix \(t\in V\) and let \(v_1,\dots,v_r\) and \(x\) be vector variables. Consider \[ f=T(v_1^b,\dots,v_r^b,t^b), \] of degree \(b\) in each \(v_i\) and degree \(0\) in \(x\). Apply \(D_{x,v_1}\) \(r\) times, then \(D_{x,v_2}\) \(r\) times, and so on up to \(v_r\). Each operator acts only on its own slot, and \(D_{x,v}^{\,r}(v^b)=\frac{b!}{m!}x^rv^m\), so \[ \begin{aligned} &D_{x,v_r}^{\,r}\cdots D_{x,v_1}^{\,r}f\\ &\quad=\Bigl(\frac{b!}{m!}\Bigr)^rT(x^rv_1^m,\dots,x^rv_r^m,t^b), \end{aligned} \] which vanishes at every point by (3), with \(Q_i=v_i^m\), and so is the zero polynomial. Before any one of these \(r^2\) operators, the degree in \(x\) is at most \(r^2-1\), while its source \(v_i\) has been differentiated at most \(r-1\) times and has degree at least \(b-r+1=m+1\ge r^2+1\). So Lemma 3.1 of A differential injectivity lemma applies before every operator, and \(f=0\).

Conclusion. So \(T(v_1^b,\dots,v_r^b,t^b)=0\) for all \(v_1,\dots,v_r,t\in V\): \(T\) vanishes on all tuples of pure powers. Pure powers span \(\operatorname{Sym}^bV\) (Lemma 1.2 of Symmetric powers and the Foulkes–Howe map), and \(T\) is multilinear, so \(T=0\). \(\square\)

2. Proof of Theorem A

For \(a\ge2\) and \(b\ge a(a-1)\), Proposition 1.1 is the vanishing statement of the diagonal criterion, Proposition 3.1 of Symmetric powers and the Foulkes–Howe map. So \(\mu_{a,b,V}\) is surjective. \(\square\)

3. An equivariant embedding

Fix a Hermitian inner product on \(V\), and give \(V^{\otimes k}\) the inner product for which \(\langle x_1\otimes\cdots\otimes x_k,y_1\otimes\cdots\otimes y_k\rangle\) equals \(\prod_i\langle x_i,y_i\rangle\). For \(g\in\mathrm{GL}(V)\) with adjoint \(g^*\), the adjoint of \(g^{\otimes k}\) is \((g^*)^{\otimes k}\), as one checks on simple tensors.

By Lemma 1.1 of Symmetric powers and the Foulkes–Howe map, applied twice, \(\operatorname{Sym}^b(\operatorname{Sym}^aV)\) is identified with a subspace of \((V^{\otimes a})^{\otimes b}\), compatibly with the actions of \(\mathrm{GL}(V)\), where \(g\) acts by \(g^{\otimes ab}\); likewise \(\operatorname{Sym}^a(\operatorname{Sym}^bV)\) with a subspace of \((V^{\otimes b})^{\otimes a}\). Give both the restricted inner products. Each subspace is invariant under \(g\) and under \(g^*\), so the adjoint of the action of \(g\) on it is the action of \(g^*\). Write \(\rho(g)\) and \(\rho'(g)\) for the two actions.

Proof of Corollary B. Let \(\mu=\mu_{a,b,V}\), which is surjective by Theorem A and equivariant (Section 2 of Symmetric powers and the Foulkes–Howe map), and let \(\mu^*:\operatorname{Sym}^a(\operatorname{Sym}^bV)\to\operatorname{Sym}^b(\operatorname{Sym}^aV)\) be its adjoint. The kernel of \(\mu^*\) is the orthogonal complement of the image of \(\mu\), which is zero, so \(\mu^*\) is injective. For \(g\in\mathrm{GL}(V)\) and all \(x,y\), \[ \begin{aligned} \langle\mu^*\rho'(g)x,y\rangle&=\langle x,\rho'(g^*)\mu y\rangle\\ &=\langle x,\mu\rho(g^*)y\rangle\\ &=\langle\rho(g)\mu^*x,y\rangle , \end{aligned} \] so \(\mu^*\rho'(g)=\rho(g)\mu^*\). \(\square\)

This argument replaces the complete reducibility of polynomial representations, used in [OpenAI-FH, Section 5], by the adjoint of the canonical map. For \(a=6\) the corollary gives an equivariant embedding \(\operatorname{Sym}^6(\operatorname{Sym}^bV)\hookrightarrow\operatorname{Sym}^b(\operatorname{Sym}^6V)\) for every \(b\ge30\).

4. Scope

The bound \(a(a-1)\) is sufficient, not claimed to be the least. Below it, surjectivity can fail. For \(a=b\) the map is an endomorphism of \(\operatorname{Sym}^a(\operatorname{Sym}^aV)\), and for \((a,b)=(5,5)\) and \((6,6)\) it has a nonzero kernel for suitable \(V\) [CIM, Theorem 6], so it is not surjective; both pairs lie below the bound of Theorem A.

5. Exercises

5.1 (easy). Write out the proof of Proposition 1.1 for \(a=2\), where \(r=1\) and \(m=b-1\).

5.2 (medium). Let \(\dim V=2\), \(b=2\), and in the basis \(e_1^2,e_1e_2,e_2^2\) of \(\operatorname{Sym}^2V\) write elements as \((A,B,C)\). Show that the nonzero symmetric bilinear form \(T\) with \(T(w,w)=B^2-4AC\) vanishes at \((u^2,u^2)\) for every \(u\in V\), but not at \((P,P)\) for every product \(P\) of two vectors. So in Proposition 1.1 products cannot be replaced by pure powers.

5.3 (easy). Explain why an endomorphism of a finite-dimensional space with a nonzero kernel is not surjective.

6. Solutions

5.1. Here \(Q=y_1\cdots y_{b-1}\). Step 1 gives \(T(xQ,tQ)=0\). Step 2 applies \(D_{y_1,t},\dots,D_{y_{b-1},t}\) to \(H=T(xQ,t^b)\); before the \(i\)-th operator the degrees are \(b-i+1\) in \(t\) and \(1\) in \(y_i\), so each is injective, and \(T(xQ,t^b)=0\). Step 3 is the case \(a=1\): \(Q\mapsto T(xQ,t^b)\) is linear and vanishes on products, so \(T(xQ_1,t^b)=0\) for all \(Q_1\in\operatorname{Sym}^{b-1}V\). Step 4 applies \(D_{x,v}\) once to \(T(v^b,t^b)\), giving \(b\,T(xv^{b-1},t^b)=0\); as \(v\) has degree \(b\ge2>0\) before it, \(T(v^b,t^b)=0\).

5.2. For \(u=se_1+te_2\), \(u^2=(s^2,2st,t^2)\) and \(B^2-4AC=4s^2t^2-4s^2t^2=0\). For \(P=(s_1e_1+t_1e_2)(s_2e_1+t_2e_2)\), with coordinates \((s_1s_2,\ s_1t_2+s_2t_1,\ t_1t_2)\), \(B^2-4AC=(s_1t_2-s_2t_1)^2\), which is nonzero for \(P=e_1e_2\). The form \(T\) is the polarization \(T(w,w')=BB'-2AC'-2A'C\).

5.3. By the rank–nullity theorem, the image has dimension \(\dim W-\dim\ker<\dim W\).

References