[proofplan] The adjointness identity is exactly the defining property of $f^\perp$ as the adjoint of multiplication by $f$. The degree assertion follows from the grading and from orthogonality of distinct homogeneous degrees: if $f$ has degree $r$ and $g$ has degree $m$, then $fg$ has degree $m+r$, so it can pair nontrivially with a homogeneous element $h$ of degree $n$ only when $m+r=n$. Nondegeneracy of the Hall [inner product](/page/Inner%20Product) on each homogeneous component then forces $f^\perp h$ to have only degree $n-r$, and no component at all if $n<r$. [/proofplan]

[step:Identify the adjointness identity from the definition of $f^\perp$] Fix $f \in \operatorname{Sym}_{\mathbb Q}$. By definition, $f^\perp:\operatorname{Sym}_{\mathbb Q}\to\operatorname{Sym}_{\mathbb Q}$ is the adjoint of the multiplication map \begin{align*} M_f:\operatorname{Sym}_{\mathbb Q} &\to \operatorname{Sym}_{\mathbb Q} \\ g &\mapsto fg. \end{align*} Therefore, for every $g,h \in \operatorname{Sym}_{\mathbb Q}$, \begin{align*} \langle M_f g,h\rangle=\langle g,f^\perp h\rangle. \end{align*} Since $M_f g=fg$, this gives \begin{align*} \langle fg,h\rangle=\langle g,f^\perp h\rangle. \end{align*} [/step]

[step:Use orthogonality of homogeneous degrees to locate $f^\perp h$]Assume now that $f \in \operatorname{Sym}_{\mathbb Q,r}$ is homogeneous of degree $r$. Fix $n \geq 0$ and $h \in \operatorname{Sym}_{\mathbb Q,n}$. Write the finite homogeneous decomposition of $f^\perp h$ as \begin{align*} f^\perp h=\sum_{k \geq 0} u_k, \end{align*} where each $u_k \in \operatorname{Sym}_{\mathbb Q,k}$ and all but finitely many $u_k$ are zero. Let $m \geq 0$ and let $g \in \operatorname{Sym}_{\mathbb Q,m}$. Since $f$ has degree $r$, the product $fg$ lies in $\operatorname{Sym}_{\mathbb Q,m+r}$. By orthogonality of distinct homogeneous components, \begin{align*} \langle fg,h\rangle=0 \end{align*} unless $m+r=n$. Using the adjointness identity already proved, \begin{align*} \langle g,f^\perp h\rangle=\langle fg,h\rangle=0 \end{align*} for every $g \in \operatorname{Sym}_{\mathbb Q,m}$ whenever $m+r \neq n$. Because $g \in \operatorname{Sym}_{\mathbb Q,m}$ pairs orthogonally with every $u_k$ for $k \neq m$, we have \begin{align*} \langle g,f^\perp h\rangle=\langle g,u_m\rangle. \end{align*} Thus, for every $m \geq 0$ with $m+r \neq n$, \begin{align*} \langle g,u_m\rangle=0 \end{align*} for all $g \in \operatorname{Sym}_{\mathbb Q,m}$. The Hall inner product is nondegenerate on $\operatorname{Sym}_{\mathbb Q,m}$, so $u_m=0$ for every $m$ with $m+r \neq n$.[/step]

[guided]We want to prove that $f^\perp h$ has a specific homogeneous degree. Since the Hall inner product is graded, the way to detect a homogeneous component is to pair against arbitrary homogeneous test elements of the same degree. Fix $m \geq 0$ and choose an arbitrary $g \in \operatorname{Sym}_{\mathbb Q,m}$. Since $f \in \operatorname{Sym}_{\mathbb Q,r}$, multiplication in the graded ring gives \begin{align*} fg \in \operatorname{Sym}_{\mathbb Q,m+r}. \end{align*} The element $h$ lies in $\operatorname{Sym}_{\mathbb Q,n}$. Distinct homogeneous components are orthogonal for the Hall inner product, so \begin{align*} \langle fg,h\rangle=0 \end{align*} whenever $m+r \neq n$. Now apply the adjointness identity from the previous step: \begin{align*} \langle g,f^\perp h\rangle=\langle fg,h\rangle. \end{align*} Therefore, whenever $m+r \neq n$, \begin{align*} \langle g,f^\perp h\rangle=0 \end{align*} for every $g \in \operatorname{Sym}_{\mathbb Q,m}$. Write the homogeneous decomposition of $f^\perp h$ as \begin{align*} f^\perp h=\sum_{k \geq 0} u_k, \end{align*} where $u_k \in \operatorname{Sym}_{\mathbb Q,k}$ and only finitely many $u_k$ are nonzero. Since $g$ has degree $m$, orthogonality of homogeneous components gives \begin{align*} \langle g,f^\perp h\rangle = \left\langle g,\sum_{k \geq 0}u_k\right\rangle = \langle g,u_m\rangle. \end{align*} Thus, for each $m$ with $m+r \neq n$, \begin{align*} \langle g,u_m\rangle=0 \end{align*} for all $g \in \operatorname{Sym}_{\mathbb Q,m}$. Nondegeneracy of the Hall inner product on $\operatorname{Sym}_{\mathbb Q,m}$ then forces $u_m=0$. Hence every homogeneous component of $f^\perp h$ vanishes except possibly the one of degree $n-r$.[/guided]

[step:Conclude the degree shift and the vanishing below degree $r$] From the previous step, the only possible nonzero homogeneous component of $f^\perp h$ has degree $m=n-r$. If $n \geq r$, this gives \begin{align*} f^\perp h \in \operatorname{Sym}_{\mathbb Q,n-r}. \end{align*} If $n<r$, then there is no integer $m \geq 0$ satisfying $m+r=n$, so every homogeneous component $u_m$ of $f^\perp h$ is zero. Hence \begin{align*} f^\perp h=0. \end{align*} Since $h \in \operatorname{Sym}_{\mathbb Q,n}$ was arbitrary, this proves \begin{align*} f^\perp\left(\operatorname{Sym}_{\mathbb Q,n}\right)\subseteq \operatorname{Sym}_{\mathbb Q,n-r}, \end{align*} with $\operatorname{Sym}_{\mathbb Q,n-r}=\{0\}$ when $n<r$. [/step]