Recall that an algebra, \(A\), over \(K\) is a vector space over \(K\) together with a multiplication operation \(A\times A\mapto A\) which is bilinear. In this section we will use the tensor product to construct the `universal’ associative algebra having an identity.

Definition A tensor of type \((r,s)\) is an element of the tensor product space \(\mathcal{T}^r_s(V)\) defined as
\begin{equation}
T^r_s(V)=\underbrace{V\otimes\dots\otimes V}_r\otimes\underbrace{V^*\otimes\dots\otimes V^*}_s.
\end{equation}
Here \(r\) is called the contravariant rank and \(s\) the covariant rank. In this context a \((0,0)\) tensor is an element of the base field \(K\), called simply a \(0\) rank tensor.

Recall that we have the following isomorphisms,
\begin{equation}
V_1^*\otimes\dots\otimes V_s^*\cong(V_1\otimes\dots\otimes V_s)^*\cong\mathcal{L}(V_1,\dots,V_r;K),
\end{equation}
so that tensors of type \((r,s)\) may be identified with multilinear functions,
\begin{equation}
f:\underbrace{V^*\times\dots\times V^*}_r\times\underbrace{V\times\dots\times V}_s\mapto K.
\end{equation}
A \(0\) rank tensor is just a scalar, the corresponding map just being scalar multiplication.

If we have another multilinear function,
\begin{equation*}
g:\underbrace{V^*\times\dots\times V^*}_p\times\underbrace{V\times\dots\times V}_q\mapto K,
\end{equation*}
which may, of course, be identified with a tensor of type \((p,q)\), then we can define a new multilinear function, such that,
\begin{equation*}
(\alpha_1,\dots,\alpha_{r+p},v_1,\dots,v_{s+q})\mapsto f(\alpha_1,\dots,\alpha_r,v_1,\dots,v_s)g(\alpha_{r+1},\dots,\alpha_{r+p},v_{s+1},\dots,v_{s+q}),
\end{equation*}
which could be identified with a tensor of type \((r+p,s+q)\). We have thus multiplied, via their respective identifications with multilinear maps, a tensor of type \((r,s)\) with a tensor of type \((p,q)\), to obtain a tensor of type \((r+p,s+q)\). The result, viewed as a multilinear map, is therefore called the tensor product, \(f\otimes g\), of the multilinear maps \(f\) and \(g\).

So defined, it is clear that this multiplication is bilinear, in the sense that,
\begin{equation}
(af_1+bf_2)\otimes g=af_1\otimes g +bf_2\otimes g,
\end{equation}
and,
\begin{equation}
f\otimes(ag_1+bg_2) =af\otimes g_1 +bf\otimes g_2,
\end{equation}
is associative but not necessarily commutative. It provides a multiplication on the space,
\begin{equation}
\mathcal{T}(V;V^*)=\bigoplus_{r,s=0}^\infty T^r_s(V).
\end{equation}
such that it becomes an algebra (here we understand \(T_0^0=K\), \(T_0^1=V\) and \(T_1^0=V^*\). This is called the tensor algebra, the name also given to the particular case of, \(\mathcal{T}(V)\), defined as
\begin{equation}
\mathcal{T}(V)=\bigoplus_{r=0}^\infty T^r(V)=K\oplus V\oplus (V\otimes V)\oplus\cdots ,
\end{equation}
equipped with the same multiplication (here \(T^r(V)=T_0^r(V)\). In fact, in this slightly simpler setting, we’ll introduce the multiplication directly without going via the identification of tensors with multilinear maps. Thus, we define the multiplication of an element of \(T^r\) with an element of \(T^s\) using the isomorphism.
\begin{equation}
(\underbrace{V\otimes\dots\otimes V}_r)\otimes(\underbrace{V\otimes\dots\otimes V}_s)\cong\underbrace{V\otimes\dots\otimes V}_{r+s},
\end{equation}
Indeed the restriction of this to \(T^r\times T^s\) provides a bilinear multiplication, \(\otimes:T^r\times T^s\mapto T^{r+s}\) (in the more general setting the only complication is that we’d need to use isomorphisms involving permutations). Equipped with this multiplication \(\mathcal{T}(V)\) is called the tensor algebra of \(V\). The tensor algebra, or rather the pair \((\mathcal{T}(V),\iota)\) where \(\iota:V\mapto T^1(V)\) is the obvious inclusion, has a universal mapping property. That is, whenever \(f:V\mapto A\) is a linear map from \(V\) into an associative algebra \(A\), with an identity, there exists a unique associative algebra homomorphism, \(F:\mathcal{T}(V)\mapto A\), with \(F(1)=1\) such that the following diagram commutes.

Here the uniqueness of \(F\) follows since \(\mathcal{T}(V)\) is generated by \(1\) and \(V\). Given then that on elements, \(v\in V\), \(F(v)=f(v)\), \(F\) is defined on the whole of \(\mathcal{T}(V)\) such that \(F(v_1\otimes\dots\otimes v_r)=f(v_1)\cdots f(v_r)\).