3
$\begingroup$

Rotman makes the following axiomatization

Definition: If $V$ is a free $k$-module of rank $n$, then a Grassmann algebra on $V$ is a $k$ algebra $G(V)$ with identity element, $e_0$, such that

  1. $G(V)$ contains $\langle e_0 \rangle \oplus V$ as a submodule, where $\langle e_0 \rangle \cong k$.
  2. $G(V)$ is generated as a $k$-algebra, by $\langle e_0 \rangle \oplus V$.
  3. $v^2=0$ for all $v \in V$.
  4. $G(V)$ is a free $k$-module of rank $2^n$.

We have Theorem 9.139, pg 747. The statement states that "the Grassmann algebra" is graded.

But the proof requires the model of a Grassman algebra constructed from part 1.

In the rest of the chapter, he also only refers to the Grassmann algebra. So question is:

Under the given axioms, are Grassmann algebras unique upto isomorphism?

In particular, why can Rotman use "the" Grassmann algebra? Or is he only using this model?

$\endgroup$
1
$\begingroup$

Yes, the Grassman algebra for $V_k$, where $V$ is an $n$ dimensional $k$ module is unique.

In brief, it can be shown that the construction of the Grassman algebra has a universal property, so that anything else satisfying the same axioms is necessarily isomorphic.

If you don't believe in the universal property of the Grassman algebra yet, but are willing to believe in the universal property of the tensor algebra $T(V)$ of $V$ over $k$, then consider this:

Suppose $V$ and $W$ are finite rank free $k$ modules. Then of course if they have different rank, $G(V)$ and $G(W)$ have to be nonisomorphic, because according to the axioms given they have different $k$ rank.

If they have the same $k$ rank, then of course they are isomorphic, say by $\theta:V\to W$. By the universal property of tensor algebras, this lifts to an isomorphism $T(V)\cong T(W)$. Then their quotients by elements of the form $x\wedge x$ are also isomorphic, but those are just $G(V)\cong G(W)$.


Also on page 750:

An astute reader will have noticed that our construction of a Grassmann algebra $G(V)$ depends not only on the free $k$-module $V$ but also on a choice of basis of $V$. Had we chosen a second basis of $V$, would the second Grassmann algebra be isomorphic to the first one?

He goes on to answer the question in

Corollary 9.142. Let $V$ be a free $k$-module, and let $B$ and $B$ be bases of $V$ . If $G(V)$ is the Grassmann algebra defined using $B$ and if $G (V)$ is the Grassmann algebra defined using $B$ , then $G(V) \cong G (V)$ as graded k-algebras.

So this, along with the fact that we shouldn't distinguish between free modules of the same rank over $k$, should convince you there is only one such algebra for a given $V_k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.