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According to the wikipedia, the exterior algebra of a $\Bbbk$-vector space $V$ is initial with respect to being unital and there existing a $\Bbbk$-linear map $j\colon V\to A$ such that $j(v)^2=0$ for all $v\in V$.

This is a reasonable algebra to consider if one is interested in measuring $k$-dimensional volumes, which are specified by $k$ linearly independent vectors, and which are degenerate if the vectors are not linearly independent (equivalently, for $char \Bbbk\neq 2$, measuring $k$-dimensional signed volumes). Then multiplication consists simply of throwing in extra vectors.

My question, inspired by the recent question on the meaning of addition in the exterior algebra, is about the meaning of co-multiplication. I happen to know virtually nothing about co-algebras outside of their formalism, so in part I am looking for answers that will help me build some intuition, geometric and otherwise, about what's happening.

  1. What is a categorical argument that the exterior algebra satisfying the above universal property has co-multiplication (and is thus a bi-algbera)?
  2. Is there a (heurisitc) geometric interpretation of the exterior algebra's co-multiplication similar to the $k$-dimensional volume tracking I sketched above?
  3. (Bonus question): Is there any geometric significance to the anti-pode that makes the exterior (bi-)algebra into a Hopf algebra?
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I cannot see what your second paragraph means, really... –  Mariano Suárez-Alvarez Jul 7 '11 at 23:39
    
Heuristically, I think of non-zero elements $v_1\wedge v_2\dots\wedge v_k$ of the exterior algebra as assigning to the (oriented) parallelepiped generated by $(v_1,\dots,v_k)$ a $k$-dimensional signed volume of $1$. Multiplication then naturally takes us from assigning $k-$ and $l$-dimensional volumes to the parallelepipeds generated by $(v_1,\dots,v_k)$ and $(w_1,\dots,w_l)$ to assigning $k+l$-dimensional signed volumes by assigning $k+l$-dimensional volume $1$ to the parallelepipied generated by $(v_1,\dots,v_k,w_1,\dots,w_l)$, unless the parallelepiped generated is of dimension $<k+l$. –  Vladimir Sotirov Jul 8 '11 at 0:45
    
if that helps you, I guess it's all great! :) In any case, not all elements of the exterior algebra are of the form $v_1\wedge\cdots\wedge v_k$. –  Mariano Suárez-Alvarez Jul 8 '11 at 3:05

2 Answers 2

up vote 7 down vote accepted

Perhaps it is first easier to understand why the symmetric algebra is a Hopf algebra. This is because $S(V)$ is nothing more than the ring of polynomial functions on $V^{\ast}$, and $V^{\ast}$ is naturally a group scheme with vector addition as the group operation. Dualizing all of the maps you get from this group structure gives you precisely the Hopf algebra structure: you get the antipode from the inverse in $V^{\ast}$ and the comultiplication from the addition in $V^{\ast}$.

The exterior algebra is a sort of "twisted" symmetric algebra, so the same thing is true for it. One way to formalize this is that the exterior algebra is precisely the symmetric algebra, but where $V$ is regarded as an odd supervector space. This is because the symmetric monoidal structure on supervector spaces is slightly different from what one might expect: it is given by $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$. See, for example, this blog post.

I think there is a physical interpretation here, but I don't know how to be precise about it. The symmetric and exterior algebras are respectively the bosonic and fermionic Fock spaces. The comultiplication ought to have a physical interpretation as "duplication of states," or something like that.


There is an idea you should understand if you haven't seen it before, and it is the notion of a group object in a category. If you write down what it means for an affine scheme $\text{Spec } R$ to be a group object in the category of affine schemes, then dualize all of the maps, you find that you have equipped $R$ with the structure of a Hopf algebra. A concise way to say this is that Hopf algebras are cogroup objects in a category of algebras, which just means that they are group objects in the opposite categories.

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Let $\Lambda V$ be the exterior algebra on $V$, and consider the tensor product algebra $\Lambda V\otimes\Lambda V$ in the sense of graded algebras, so that if $a$, $b$, $c$ and $d$ are homogeneous elements in $\Lambda V$ we have $$a\otimes b\cdot c\otimes d=(-1)^{|b||c|}ac\otimes bd.$$ If $v\in V$, then you can chech that $(1\otimes v+v\otimes1)^2=0$. Therefore we have a map $$j:v\in V\mapsto v\otimes1+1\otimes v\in\Lambda V\otimes\Lambda V$$ such that $j(v)^2=0$ for all $v\in V$. As you observed, this $j$ then induces an algebra map $$\Delta:\Lambda V\to\Lambda V\otimes\Lambda V.$$ Using the universal property of $\Lambda V$ it is easy to show that this turns $\Lambda V$ into a Hopf algebra in the graded sense. For example, we need to show $\Delta$ is coassociative: but the maps $\Delta\otimes1_V\circ\Delta$ and $1_\V\otimes\Delta\circ\Delta$ are two algebra maps $\Lambda V\to\Lambda V\otimes\Lambda V\otimes\Lambda V$, so the universal property tells us that to check they coincide it is enough to show their restrictions to $V\subseteq\Lambda V$ coincide, and this can be done by computing explicitely.

It is important that the tensor product algebra $\Lambda V\otimes\Lambda V$ be taken in the graded sense here. In the ungraded sense, $\Lambda V$ is generally $not$ a Hopf algebra: for example, when $V$ is one dimensional, $\Lambda V\cong k[x]/(x^2)$ as an algebra and if the characteristic of the field $k$ is not two, this cannot be made into a Hopf algebra in any way.

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