A tensor, \(T\in T^r(V)\), is called skew-symmetric if \(P_\sigma(T)=\sgn(\sigma)T\) for all \(\sigma\in S_r\). The subspace in \(T^r(V)\) of all skew-symmetric tensors will be denoted \(\Lambda^r(V)\).
Define on \(T^r(V)\) the linear operator,
\begin{equation}
A=\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\sigma)P_\sigma.
\end{equation}
This is called the antisymmetrization on \(T^r(V)\). For any \(T\in T^r(V)\), \(A(T)\) is skew-symmetric, since for any \(\tau\in S_r\),
\begin{align*}
P_\tau\left(\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\sigma)P_\sigma(T)\right)&=\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\sigma)P_{\tau\sigma}(T)\\
&=\sgn(\tau)\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\tau\sigma)P_{\tau\sigma}(T)\\
&=\sgn(\tau)\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\sigma)P_{\sigma}(T)\\
&=\sgn(\tau)A(T).
\end{align*}
Conversely, suppose \(T\) is skew-symmetric, then
\begin{equation}
A(T)=\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\sigma)P_\sigma(T)=\frac{1}{r!}\sum_{\sigma\in S_r}\sgn(\sigma)^2T=T,
\end{equation}
so that \(\img A=\Lambda^r(V)\) and \(A^2=A\), so that \(A\) is a projector onto \(\Lambda^r(V)\).
If \(\{e_i\}\) is a basis for 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}
A(e_{i_1}\otimes\dots\otimes e_{i_r})=e_{i_1}\wedge\dots\wedge e_{i_r}.
\end{equation}
The symbol \(\wedge\) is called the exterior or wedge product. Since by definition, two pure tensors,
\begin{equation*}
e_{i_1}\otimes\dots\otimes e_{i_j}\otimes\dots\otimes e_{i_k}\otimes\dots\otimes e_{i_r},
\end{equation*}
and
\begin{equation*}
e_{i_1}\otimes\dots\otimes e_{i_k}\otimes\dots\otimes e_{i_j}\otimes\dots\otimes e_{i_r},
\end{equation*}
differing only by the interchange of the pair \(e_{i_j}\) and \(e_{i_k}\), are related by a permutation \(\sigma\) with \(\sgn(\sigma)=-1\), we have,
\begin{equation*}
e_{i_1}\wedge\dots\wedge e_{i_j}\wedge\dots\wedge e_{i_k}\wedge\dots\wedge e_{i_r}=-e_{i_1}\wedge\dots\wedge e_{i_k}\wedge\dots\wedge e_{i_j}\wedge\dots\wedge e_{i_r}.
\end{equation*}
In particular, if \(i_j=i_k\) for some \(j\neq k\), then \(e_{i_1}\wedge\dots\wedge e_{i_r}=0\). It also follows that \(\Lambda^r(V)\) is spanned by tensors of the form, \(e_{i_1}\wedge\dots\wedge e_{i_r}\), such that, \(1\leq i_1
\begin{equation}
\dim\Lambda^r(V)={n\choose r},
\end{equation}
with \(\dim\Lambda^n(V)=1\). We define,
\begin{equation}
\Lambda(V)=\bigoplus_{r=0}^n\Lambda^r(V),
\end{equation}
(\(\dim\Lambda(V)=2^n\)) and introduce a multiplication according to \(T_1\wedge T_2=A(T_1\otimes T_2)\) for any \(T_1\in\Lambda^r(V)\) and \(T_2\in\Lambda^s(V)\). Then for any, \(T_1\in T^r(V)\), \(T_2\in T^r(V)\) and \(T_3\in T^s(V)\),
\begin{align*}
(T_1\wedge T_2)\wedge T_3&=A(A(T_1\otimes T_2)\otimes T_3)\\
&=A\left(\frac{1}{(r+s)!}\sum_{\sigma\in S_{r+s}}\sgn(\sigma)P_\sigma(T_1\otimes T_2)\otimes T_3\right)\\
&=\frac{1}{(r+s)!}\sum_{\sigma\in S_{r+s}}\sgn(\sigma)A(P_\sigma(T_1\otimes T_2)\otimes T_3)\\
&=\frac{1}{(r+s)!}\sum_{\sigma\in S_{r+s}}\sgn(\sigma)^2A(T_1\otimes T_2\otimes T_3)\\
&=A(T_1\otimes T_2\otimes T_3),
\end{align*}
and similarly for \(T_1\wedge(T_2\wedge T_3)=A(T_1\otimes T_2\otimes T_3)\). We conclude, therefore, that the wedge product is associative. Also, for any \(T_1\in T^r(V)\) and \(T_2\in T^s(V)\), we have, \(A(T_1\otimes T_2)=(-1)^{rs}A(T_2\otimes T_1)\) so, in particular, \(T_1\wedge T_2=(-1)^{rs}T_2\wedge T_1\), for any \(T_1\in\Lambda^r(V)\) and \(T_2\in\Lambda^s(V)\).
As with the symmetric algebra, let us now realise \(\Lambda(V)\) as a quotient of the tensor algebra \(T(V)\).
Definition The exterior algebra on the vector space \(V\) over the field \(K\) is the quotient \(T(V)/J\) of the tensor algebra \(T(V)\) by the ideal \(J\) generated by the elements \(v\otimes v\) for all \(v\in V\).
As in the symmetric algebra case, we define \(J^r=T^r(V)\cap J\) so that \(J=\bigoplus_{r=0}^\infty J^r\). Then defining, \(\tilde{\Lambda}(V)=T(V)/J\), it follows as before that, \(\tilde{\Lambda}(V)=\bigoplus_{r=0}^\infty T^r(V)/J^r\). Thus we define, \(\tilde{\Lambda}^r(V)=T^r(V)/J^r\), and seek to relate this to \(\Lambda^r(V)\) defined above.
An alternative definition of the ideal \(J\), is as the ideal generated by the elements, \(u\otimes v+v\otimes u\), for any \(u,v\in V\). The equivalence of these definitions amounts to observing that for any \(u,v\in V\),
\begin{equation*}
(u+v)\otimes(u+v)-u\otimes u-v\otimes v=u\otimes v+v\otimes u.
\end{equation*}
Then, by an argument similar to the one we used in the symmetric case, this ideal is equivalent to the ideal generated by \(T-\sgn(\sigma)P_\sigma(T)\) for all \(T\in T^r(V)\) and any \(\sigma\in S_r\). Once again abusing notation, we’ll denote the product of two elements of \(\tilde{\Lambda}(V)\), \((T_1+J)(T_2+J)\), as \(T_1\wedge T_2\) rather than \(T_1\otimes T_2+J\). Thus the image in, \(\tilde{\Lambda}(V)\), of some pure tensor, \(v_1\otimes\dots\otimes v_r\in T^r(V)\), is denoted \(v_1\wedge\dots\wedge v_r\). Then since, \(T-\sgn(\sigma)P_\sigma(T)\in J\), it follows that, \(T_1\wedge T_2=(-1)^{rs}T_2\wedge T_1\), for any \(T_1\in\tilde{\Lambda}^r(V)\) and \(T_2\in\tilde{\Lambda}^s(V)\).
Just as in the symmetric case, skew-symmetric tensors and the exterior algebra inherit universal properties from the tensor product and tensor algebra respectively. The proofs follow those of the symmetric case.
Proposition If \(\iota\) is the \(r\)-linear function, \(\iota:V\times\dots\times V\mapto\tilde{\Lambda}^r(V)\), defined as , \(\iota(v_1,\dots,v_r)=v_1\cdots v_r\), then \((\tilde{\Lambda}^r(V),\iota)\) has the following universal mapping property: whenever \(f:V\times\dots\times V\mapto W\) is an alternating\footnote{We already met alternating forms in our earlier discussion of determinants, the only generalisation here is that the target space is another vector space.} \(r\)-linear function with values in a vector space \(W\) there exists a unique linear map \(L:\tilde{\Lambda}^r(V)\mapto W\) such that \(f=L\iota\).
Consequently, the space of linear maps \(\mathcal{L}(\tilde{\Lambda}^r(V),W)\) is isomorphic to the vector space of alternating \(r\)-linear functions from \(V\times\dots\times V\) to \(W\) and in particular \(\tilde{\Lambda}^r(V)^*\), the dual space of \(\tilde{\Lambda}^r(V)\), is isomorphic to the space of all alternating \(r\)-linear forms on \(V\times\dots\times V\).
Given a basis, \(\{e_i\}\), of \(V\), the pure tensors, \(e_{i_1}\otimes\dots\otimes e_{i_r}\), form a basis of \(T^r(V)\), and so the, \(e_{i_1}\wedge\dots\wedge e_{i_r}\), span \(\tilde{\Lambda}^r(V)\). In fact its clear that this space is already spanned by the set \(e_{i_1}\wedge\dots\wedge e_{i_r}\) with \(1\leq i_1
\begin{equation}
T^2(V)=S^2(V)\oplus\Lambda^2(V).
\end{equation}
Remark The elements of \(\Lambda^r(V)\) are called \(r\)-vectors. An \(r\)-vector which can be written \(v_1\wedge\dots\wedge v_r\) for some \(v_i\in V\) will be called a pure \(r\)-vector.