I have several questions on the exterior algebra of a vector space:

Q1:When has the exterior algebra A (viewed just as an algebra, not considered as a graded algebra) of an $n$-dimensional vector space over a field $K$ the structure of a Hopf algebra? (depending on n and K)

Note that it is not always a Hopf algebra, for example in the easiest case the exterior algebra is $K[x]/(x²)$ and this should be a Hopf algebra iff the characteristic of the field is 2.

Q2: Is there a finite dimensional, nonprojective module M over A with $Ext_A^{1}(M,M)=0$?

This question has the answer no when the exterior algebra is a Hopf algebra and thus this question is related to Q1.

(Q2 is also open in the graded case and has a positive solution in a special case, see the last chapter of https://arxiv.org/pdf/1701.01149.pdf )

Q3:Can one classify all periodic modules over this algebra?

In general the exterior is a wild algebra for more than 2 variables and it is hopeless to give a classification of all indecomposable modules, but maybe there is an interesting classification of special modules such as indecomposable periodic modules (a module is periodic in case $\Omega^n(M) \cong M$ for some $n$).

  • $\begingroup$ It's hard to make the question meaningful... What is the coproduct? $\endgroup$
    – YCor
    Nov 20, 2016 at 8:37
  • $\begingroup$ What do you mean? The question just is: When is this given algebra a Hopf algebra (so you are allowed to choose the coproduct making it a hopf algebra, if it exists) $\endgroup$
    – Mare
    Nov 20, 2016 at 8:39
  • 2
    $\begingroup$ An exterior algebra is a Hopf algebra, for the comultiplication $x\mapsto x\otimes 1+1\otimes x$ and the antipode $x\mapsto -x$. $\endgroup$
    – abx
    Nov 20, 2016 at 10:54
  • 3
    $\begingroup$ @abx: See the answer of Mariano Suárez-Álvarez in math.stackexchange.com/questions/50187/… $\endgroup$
    – Mare
    Nov 20, 2016 at 11:43
  • 1
    $\begingroup$ You may have noticed that your question is still unclear. For example, if by exterior algebra you mean the graded algebra given by the direct sum of the exterior powers of the given vector space, then it is of course a Hopf algebra as the comment of abx shows, and it is also isomorphic to its dual. Note that the key word is 'graded' $\endgroup$
    – F Zaldivar
    Nov 20, 2016 at 14:19

2 Answers 2


I will show that the exterior algebra over a nonzero vector space can only support a co-algebra structure in characteristic $2$. Over a field of characteristic $2$, the map $v \mapsto v \otimes 1 + 1 \otimes v$ is a coproduct which, with the identity map as antipode, makes $\bigwedge^{\bullet} V$ into a Hopf algebra, so the answer is "if and only if the ground field has characteristic $2$.

Notation Let $\Lambda$ be the exterior algebra on $V$, and let $M$ be its radical, $\bigoplus_{j \geq 1} \bigwedge^j V$. By a coproduct, we mean an algebra map $\Delta: \Lambda \to \Lambda \otimes \Lambda$ with the standard ring structure $(u_1 \otimes v_1) (u_2 \otimes v_2) = (u_1 u_2) \otimes (v_1 v_2)$ on $\Lambda \otimes \Lambda$. In the comments, people discuss more exotic ring structures, but the OP makes it clear that is not what they are looking for. Suppose that $\Delta$ is a coproduct and $\epsilon: \Lambda \to k$ is a counit, making a coalgebra structure.

Since $\epsilon$ is a map of $k$-algebras, its kernel must be $M$ and it must be the obvious map $\Lambda \to \Lambda/M \cong k$. Let $x \in V$. The counit axiom is equivalent to saying that $\Delta(x) \equiv x \otimes 1 \bmod \Lambda \otimes M$, and $\Delta(x) \equiv 1 \otimes x \bmod M \otimes \Lambda$, so $$\Delta(x) \equiv x \otimes 1 + 1 \otimes x \bmod M \otimes M.$$

We deduce that $$\Delta(x^2) \equiv x^2 \otimes 1 + 2 x \otimes x + 1 \otimes x^2 \bmod M^2 \otimes M + M \otimes M^2 .$$ But $x^2=0$, so this says $$2 x \otimes x \equiv 0 \bmod M^2 \otimes M + M \otimes M^2.$$ If $2 \neq 0$ and $x \neq 0$, this is a contradiction. So there are no solutions over nonzero vector spaces in characteristic not $2$.

  • $\begingroup$ Cool proof. This also shows that in case the exterior algebra is a Hopf algebra, then it has to be symmetric. It seems unknown whether every local Hopf algebra is symmetric, see mathoverflow.net/questions/202339/… . $\endgroup$
    – Mare
    Nov 30, 2018 at 8:08

Another proof: The exterior algebra is Koszul, it's Koszul dual is the symmetric algebra, that is commutative, but NOT super commutative, unless Char=2. Assuming $char \neq 2$, the exterior algebra cannot be Hopf because for $H$ a Hopf algebra, $Ext_H(k,k)$ is a subalgebra of Hochschild cohomology of H, hence, it should be super commutative.


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