2
$\begingroup$

There are basically two different definitions of the exterior algebra and exterior powers of a vector space $V$. I here want to concentrate on the one where we define the exterior algebra as a quotient of the tensor algebra. Now for this, i have seen two different definitions: In some books the ideal generated by all tensors of the form $v \otimes v, v \in V$ is factorized out. In other books(e.g. Kostrikin,Manin: Linear Algebra and Geometry) the ideal generated by all tensors of the form $t- sgn(\sigma)\phi_\sigma(t), t \in \underbrace{V \otimes ... \otimes V}_{k}=:T^k(V), \sigma \in S_k, k=1,2,3,...$, where the symmetric group $S_k$ is acting on tensors in the natural way (denoted by $\phi_\sigma$). Or formulated in terms of the $k$-th homogeous part of the respective ideal: the former corresponds to factorizing out the subspace of $T^k(V)$, spanned by all $v_1 \otimes ... \otimes v_k$ with $v_i = v_{i+1}$ (or equivalently with $v_i = v_j$ for some $i \neq j$). The latter corresponds to factorizing out the subspace generated by $v_1 \otimes ... \otimes v_k - sgn(\sigma)v_{\sigma(1)} \otimes ... \otimes v_{\sigma(k)}$. So far, is this correct? My confusion now is the following: One could formulate this also in terms of the respective canonical maps $\lambda_k$ from $V^k$ to the respective quotient of $T^k(V)$. In the first case, the definition says that $\lambda_k$ is an alternating map, while in the second case, the definition says that it is a skew-symmetric map. Now, an alternating map is always skew-symmetric but the converse is only true in characteristic $\neq 2$. Why then do i find these two different definitions which are supposed to be used in all characteristics? Note: i do not ask about the definition one often finds in differential-geometric contexts, where one defines the $k$-th exterior power as the image of the alternation-operator (and which needs division by $k!$). Thanks.

$\endgroup$

1 Answer 1

2
$\begingroup$

The second definition is just wrong in characteristic $2$. Any author who uses it is either implicitly ignoring the characteristic $2$ case or is being imprecise. You find similar problems with the definition of Clifford algebras; there are again two variants and one of them is wrong in characteristic $2$.

$\endgroup$
1
  • $\begingroup$ ok, i checked again all sources i could find and where i found the supposedly wrong definition is the wiki of wolfram and the mentioned book by Kostrikin and Manin which interestingly thoroughly discusses also the definition via the alternation operator and then introduces the definition via quotient by remarking that the other definition has the drawback of requiring division by factorials. $\endgroup$
    – Mekanik
    Jan 8, 2015 at 19:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .