In this and the next section we will identify symmetric and skew-symmetric tensors within \(T(V)\), and demonstrate, that with a suitably defined multiplication, they form subalgebras of \(T(V)\). In both cases we’ll then realise these algebras as quotients of \(T(V)\).

To any permutation \(\sigma\in S_r\) denote by \(P_\sigma:T^r(V)\mapto T^r(V)\) the linear operator defined on pure tensors by \(P_\sigma(v_1\otimes\dots\otimes v_r)=v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(r)}\). A tensor, \(T\in T^r(V)\), is called symmetric if \(P_\sigma(T)=T\) for all \(\sigma\in S_r\). The subspace in \(T^r(V)\) of all symmetric tensors will be denoted \(S^r(V)\).

Consider the linear operator, \(S\), on \(T^r(V)\), defined as,
\begin{equation}
S=\frac{1}{r!}\sum_{\sigma\in S_r}P_\sigma.
\end{equation}
This is called the symmetrization on \(T^r(V)\). For any permutation, \(\sigma\in S_r\), \(P_\sigma S=S\), so for any \(T\in T^r(V)\), \(S(T)\) is symmetric. Conversely, it is clear that if \(T\) is symmetric, then \(S(T)=T\). Thus \(\img S=S^r(V)\) and \(S^2=S\), so \(S\) is a projector onto \(S^r(V)\).

Example Consider a 2-dimensional vector space over \(\CC\) with basis \(\{e_1,e_2\}\). Then on the natural basis of \(T^3(V)\), we have
\begin{align*}
S(e_1\otimes e_1\otimes e_1)&=e_1\otimes e_1\otimes e_1\\
S(e_1\otimes e_1\otimes e_2)&=\frac{1}{3}(e_1\otimes e_1\otimes e_2+e_1\otimes e_2\otimes e_1+e_2\otimes e_1\otimes e_1)\\
S(e_1\otimes e_2\otimes e_1)&=\frac{1}{3}(e_1\otimes e_1\otimes e_2+e_1\otimes e_2\otimes e_1+e_2\otimes e_1\otimes e_1)\\
S(e_2\otimes e_1\otimes e_1)&=\frac{1}{3}(e_1\otimes e_1\otimes e_2+e_1\otimes e_2\otimes e_1+e_2\otimes e_1\otimes e_1)\\
S(e_1\otimes e_2\otimes e_2)&=\frac{1}{3}(e_1\otimes e_2\otimes e_2+e_2\otimes e_1\otimes e_2+e_2\otimes e_2\otimes e_1)\\
S(e_2\otimes e_1\otimes e_1)&=\frac{1}{3}(e_1\otimes e_2\otimes e_2+e_2\otimes e_1\otimes e_2+e_2\otimes e_2\otimes e_1)\\
S(e_2\otimes e_2\otimes e_1)&=\frac{1}{3}(e_1\otimes e_2\otimes e_2+e_2\otimes e_1\otimes e_2+e_2\otimes e_2\otimes e_1)\\
S(e_2\otimes e_2\otimes e_2)&=e_2\otimes e_2\otimes e_2
\end{align*}

Let us consider the dimension of \(S^r(V)\). If \(\{e_i\}\) is a basis of the \(n\)-dimensional vector space, \(V\), then all pure tensors of the form, \(e_{i_1}\otimes\dots\otimes e_{i_r}\), form a basis of \(T^r(V)\). A standard notation is to write,
\begin{equation}
S(e_{i_1}\otimes\dots\otimes e_{i_r})=e_{i_1}\dots e_{i_r}.
\end{equation}
The tensors, \(e_{i_1}\dots e_{i_r}\), clearly span \(S^r(V)\), but, as the Example makes clear, whenever \(\{i_1,\dots,i_r\}\) and \(\{j_1,\dots,j_r\}\) are identical as sets \(e_{i_1}\dots e_{i_r}=e_{j_1}\dots e_{j_r}\). In other words, \(e_{i_1}\dots e_{i_r}\) only depends on the number of times each \(e_i\) appears in the product, so we can write \(e_{i_1}\dots e_{i_r}=e_1^{a_1}\dots e_n^{a_n}\) where \(a_i\) is the multiplicity of \(e_i\) in \(e_{i_1}\dots e_{i_r}\), and \(a_1+\dots+a_n=r\). It is then clear that the tensors, \(e_1^{a_1}\dots e_n^{a_n}\), are linearly independent — for distinct \(n\)-tuples, \((a_1,\dots,a_n)\) and \((b_1,\dots, b_n)\), \(e_1^{a_1}\dots e_n^{a_n}\) and \(e_1^{b_1}\dots e_n^{b_n}\) are linear combinations of non-intersecting subsets of basis elements of \(T^r(V)\). Thus the \(e_1^{a_1}\dots e_n^{a_n}\) are a basis for \(S^r(V)\) and so to determine the dimension of \(S^r(V)\), we must count the number of distinct \(n\)-tuples \((a_1,\dots,a_n)\), \(a_i\in\ZZ_{\geq0}\), such that \(a_1+\dots+a_n=r\). A nice way of understanding this counting problem is through Feller’s `stars and bars’. Suppose \(r=8\) and \(n=5\) so that we wish to determine the dimension of the space \(S^8(V)\) where \(V\) is a \(5\)-dimensional vector space. Then each valid \(5\)-tuple corresponds to a diagram such as,
\begin{equation*}
|\star\star||\star\star|\star\star\star|\star|,
\end{equation*}
in which, reading left to right, the number of stars between the \(i\)th pair of bars corresponds to \(a_i\), so in particular, this example corresponds to \((2,0,2,3,1)\). We therefore need to count the number of possible arrangements of \(n-1\) bars and \(r\) stars, or in other words, the number of ways of choosing \(r\) star locations from the \(n+r-1\) possible locations. Thus we have that for an \(n\)-dimensional vector space \(V\),
\begin{equation}
\dim S^r(V)={n+r-1\choose r}.
\end{equation}

On the space, \(S(V)=\bigoplus_{r=0}^\infty S^r(V)\), we can define a multiplication according to, \(T_1\cdot T_2=S(T_1\otimes T_2)\), for any \(T_1\in S^r(V)\) and \(T_2\in S^s(V)\). Equipped with this multiplication \(S(V)\) becomes a commutative associative algebra. That the product is commutative is clear. That it is associative follows since, for any \(T_1\in S^r(V)\), \(T_2\in S^s(V)\) and \(T_3\in S^t(V)\),
\begin{align*}
(T_1\cdot T_2)\cdot T_3&=S(S(T_1\otimes T_2)\otimes T_3)\\
&=S\left(\frac{1}{(r+s)!}\sum_{\sigma\in S_{r+s}}P_\sigma(T_1\otimes T_2)\otimes T_3\right)\\
&=\frac{1}{(r+s)!}\sum_{\sigma\in S_{r+s}}S(P_\sigma(T_1\otimes T_2)\otimes T_3)\\
&=S(T_1\otimes T_2\otimes T_3),
\end{align*}
and similarly for \(T_1\cdot(T_2\cdot T_3)=S(T_1\otimes T_2\otimes T_3)\).

As already discussed, the tensor product provides a multiplication on the space \(T(V)=\bigoplus_{r=0}^\infty T^r(V)\) such that it becomes an associative algebra with identity. Moreover, by virtue of its universal mapping property we should expect to be able to realise the algebra \(S(V)\) as a quotient of \(T(V)\) by a certain ideal\footnote{Recall subspace \(I\) of an algebra \(A\) is an ideal of \(A\) if for all \(a\in A\) and all \(i\in I\), \(ai\in I\) and \(ia\in I\). In this case the space \(A/I\) is an algebra with the multiplication inherited from \(A\) according to \((a+I)\cdot(b+I)=ab+I\)}, \(I\). Indeed, if \(\pi:T(V)\mapto T(V)/I\) is the quotient map, then for \(u,v\in V\) we’ll want that \(\pi(u\otimes v)=\pi(v\otimes u)\) in \(T(V)/I\), that is, we’ll want \(u\otimes v-v\otimes u\in I\). We are led, therefore, to the following definition.

Definition The symmetric algebra on the vector space \(V\) over the field \(K\) is the quotient \(T(V)/I\) of the tensor algebra \(T(V)\) by the ideal \(I\) generated by the elements \(u\otimes v-v\otimes u\) for all \(u,v\in V\).

Let us denote by \(\tilde{S}(V)\), the symmetric algebra as defined here. Then defining \(I^r=T^r(V)\cap I\), it’s not difficult to see that \(I=\bigoplus_{r=0}^\infty I^r\). In fact, \(\tilde{S}(V)=\bigoplus_{r=0}^\infty T^r(V)/I^r\), which we can see by observing that the linear map defined as \(\sum_r(T_r+I^r)\mapsto \sum_rT_r+I\), where \(T_r\in T^r(V)\), is clearly surjective and is injective since if \(\sum_rT^r\in I\) then \(T^r\in I^r\). Thus, setting, \(\tilde{S}^r(V)=T^r(V)/I^r\), so that \(\tilde{S}(V)=\bigoplus_{r=0}^\infty\tilde{S}^r(V)\), we will want to establish that \(\tilde{S}^r(V)\) and \(S^r(V)\) are isomorphic, from which, it will immediately follow that \(\tilde{S}(V)\) and \(S(V)\) are isomorphic.

There is an alternative description of the ideal, \(I\). Let us denote by, \(I’\), the ideal generated by all elements, \(T-P_\sigma(T)\), for any tensor, \(T\in T^r(V)\), and any permutation, \(\sigma\in S_r\). Now for any pure tensor, \(v_1\otimes\dots\otimes v_r\),
\begin{equation*}
v_1\otimes\dots\otimes v_r-v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(r)},
\end{equation*}
can be written as a sum of terms of the form
\begin{equation*}
v_1\otimes\dots\otimes v_{i_1}\otimes v_{i_1+1}\otimes\dots\otimes v_r-v_1\otimes\dots\otimes v_{i_1+1}\otimes v_{i_1}\otimes\dots\otimes v_r,\end{equation*}
in which only neighbouring factors are transposed. Since each of these clearly belongs to the ideal, \(I\), as originally defined, it follows that \(I’\subseteq I\). The reverse inclusion is obvious so we have \(I’=I\).
In particular, \(T(V)/I\) is commutative since for any pure tensors, \(T_1,T_2\in T(V)\), \(T_2\otimes T_1=P_\sigma(T_1\otimes T_2)\), for some permutation, \(\sigma\), so that, \(T_1\otimes T_2-T_2\otimes T_1=T_1\otimes T_2-P_\sigma(T_1\otimes T_2)\in I\). That is, \((T_1+I)(T_2+I)=T_1\otimes T_2+I=T_2\otimes T_1+I=(T_2+I)(T_1+I)\).

Abusing notation, for any \(v_1\otimes\dots\otimes v_r\in T^r(V)\), \(v_i\in V\), let us write its image in \(\tilde{S}^r(V)\), via the quotient map, \(\pi\), as \(v_1\cdots v_r\). Recall that the tensor product was defined via a universal mapping property. In particular, whenever we have an \(r\)-linear function \(f:V\times\dots\times V\mapto W\), where \(W\) is some vector space, then there is a unique linear mapping \(L:T^r(V)\mapto W\) such that \(f=L\iota\) where \(\iota \) was the \(r\)-linear function \(\iota(v_1,\dots,v_r)=v_1\otimes\dots\otimes v_r\). This leads to the following result for \(\tilde{S}^r(V)\).

Proposition If \(\iota\) is the \(r\)-linear function, \(\iota:V\times\dots\times V\mapto\tilde{S}^r(V)\), defined as, \(\iota(v_1,\dots,v_r)=v_1\cdots v_r\), then \((\tilde{S}^r(V),\iota)\) has the following universal mapping property: whenever \(f:V\times\dots\times V\mapto W\) is a symmetric \(r\)-linear function with values in a vector space \(W\) there exists a unique linear map \(L:\tilde{S}^r(V)\mapto W\) such that \(f=L\iota\).

Proof From the universal mapping property of the tensor product we have a map \(L’:T^r(V)\mapto W\) such that on pure tensors \(L'(v_1\otimes\dots\otimes v_r)=f(v_1,\dots,v_r)\). But since \(f(v_1,\dots,v_i,v_{i+1},\dots,v_r)=f(v_1,\dots,v_{i+1},v_i,\dots,v_r)\) it is clear that for any \(T\in I^r\), \(L'(T)=0\), and so \(L’\) factorises as, \(L’=L\pi\), where \(\pi\) is the quotient map \(\pi:T^r(V)\mapto T^r(V)/I^r\) and \(L:T^r(V)/I^r\mapto W\) is desired linear map.\(\blacksquare\)

As a consequence, we have that the space of linear maps \(\mathcal{L}(\tilde{S}^r(V),W)\) is isomorphic to the vector space of symmetric \(r\)-linear functions from \(V\times\dots\times V\) to \(W\) and in particular that \(\tilde{S}^r(V)^*\), the dual space of \(\tilde{S}^r(V)\), is isomorphic to the space of all symmetric \(r\)-linear forms on \(V\times\dots\times V\).

Recall that the tensor algebra, \(T(V)\), has a universal mapping property whereby whenever \(f:V\mapto A\) is a linear map from \(V\) into an associative algebra \(A\) with identity there exists a unique algebra homomorphism, \(F:T(V)\mapto A\), with \(F(1)=1\) and such that \(F(v)=f(v)\) with \(F(v_1\otimes\dots\otimes v_r)=f(v_1)\cdots f(v_r)\). This leads to the following result for \(\tilde{S}(V)\).

Proposition If \(\iota\) is the linear map embedding \(V\) in \(T(V)\) then \((\tilde{S}(V),\iota)\) has the following universal mapping property: whenever \(f:V\mapto A\) is a linear map from \(V\) into a commutative associative algebra \(A\) with identity, there exists a unique algebra homomorphism, \(F:\tilde{S}(V)\mapto A\), with \(F(1)=1\) such that \(F(v)=f(v)\) with \(F(v_1\cdots v_r)=f(v_1)\cdots f(v_r)\).

Proof From the universal mapping property of the tensor algebra we have an algebra homomorphism \(F’:T(V)\mapto A\) such that \(F'(v)=f(v)\) and since \(A\) is commutative we have \(F'(u\otimes v-v\otimes u)=0\), so \(I\in\ker F’\) and \(F’\) factorises as \(F’=F\pi\) where \(\pi\) is the quotient map \(\pi:T(V)\mapto T(V)/I\) and \(F:T(V)/I\mapto A\) is the desired algebra homomorphism.\(\blacksquare\)

Now if \(\{e_i\}\) is a basis of \(V\), then \(r\)-fold (tensor) products of the \(e_i\), \(e_{i_1}\otimes\cdots\otimes e_{i_r}\), span \(T^r(V)\). But since \(T(V)/I\) is commutative, this means that the elements, \(e_1^{a_1}\cdots e_n^{a_n}\), such that \(a_1+\dots+a_n=r\), must span \(\tilde{S}^r(V)\). Now recall that \(S^r(V)\) was defined as the image of the symmetrization operator, \(S\), on \(T^r(V)\), and that \(S\) is a projector. This means that \(T^r(V)=\ker S\oplus\img S=\ker S\oplus S^r(V)\). Clearly any element of \(I^r\) belongs to \(\ker S\), so \(I^r\subseteq\ker S\). But if there was some \(T\in\ker S\) such that \(T\notin I^r\) then \(\pi(T)\neq0\) and we must be able to express \(\pi(T)\) as a linear combination of the \(e_1^{a_1}\cdots e_n^{a_n}\). Thus, using these same linear coefficients, we can chose a tensor, \(T’\in T^r(V)\), as a linear combination of pure tensors of the form,
\begin{equation*}
\underbrace{e_1\otimes\dots\otimes e_1}_{a_1}\otimes\dots\otimes\underbrace{e_n\otimes\dots\otimes e_n}_{a_n},
\end{equation*}
each tensor in this linear combination corresponding to a distinct \(n\)-tuple, \((a_1,\dots,a_n)\), such that \(\pi(T)=\pi(T’)\). Then, \(T-T’\in I^r\), so \(S(T)=S(T’)\).
But \(S(T’)\) cannot be zero, since the symmetrization of distinct pure tensors in the linear combination, \(T’\), are non-zero linear combinations of distinct sets of basis elements of \(T^r(V)\). Thus, \(S(T)\neq0\), contradicting our initial assumption. It follows that \(\ker S=I^r\) and we have established that \begin{equation}
T^r(V)=S^r(V)\oplus I^r.
\end{equation}
In particular, this means that \(\dim T^r(V)/I^r=\dim S^r(V)\), so that the elements, \(e_1^{a_1}\cdots e_n^{a_n}\), such that \(a_1+\dots+a_n=r\), are a basis for \(\tilde{S}^r(V)\) and of course that \(\tilde{S}^r(V)\cong S^r(V)\) with this isomorphism such that, \(T+\ker S\mapsto S(T)\), mapping basis elements whose identification has already been anticipated by our abuse of notation. This clearly extends to a (grade preserving) algebra isomorphism \(\tilde{S}(V)\cong S(V)\).