Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be a field, let $V$ and $W$ be $k$-vector spaces of dimensions $n$ and $m$ respectively, and let $f:V\to W$ be a $k$-linear transformation. Let $\Lambda(V)$ and $\Lambda(W)$ denote the exterior algebras of $V$ and $W$ respectively. So we have $$\Lambda(V) = \Lambda^0(V)\oplus\Lambda^1(V)\oplus\cdots\oplus\Lambda^n(V)$$ and $$\Lambda(W) = \Lambda^0(W)\oplus\Lambda^1(W)\oplus\cdots\oplus\Lambda^m(W).$$

The wikipedia page on exterior algebras states that there is a unique function $\Lambda(f):\Lambda(V)\to\Lambda(W)$ such that $\Lambda(f)|_{\Lambda^1(V)}:\Lambda^1(V)\to\Lambda^1(W)$ is defined by $\Lambda(f)(v)=f(v)$.

In fact, $\Lambda(f)$ preserves grading (i.e. it can be written as a sum of maps $\Lambda^k(f):=\Lambda(f)|_{\Lambda^k(V)}:\Lambda^k(V)\to\Lambda^k(W)$). If $1\leq k \leq n$, then $\Lambda(f)$ is given by $$\Lambda^k(f)(v_1\wedge\cdots\wedge v_k) = f(v_1)\wedge\cdots\wedge f(v_k).$$

I do not understand how this function acts on $\Lambda^0(V)=k$. I know that we have a map $$\Lambda^0(f):\Lambda^0(V)\to\Lambda^0(W)$$ which is really the same as $$\Lambda^0(f):k\to k.$$

My question has two parts: what is $\Lambda^0(f)$ and how is it determined from the universal mapping property for exterior algebras?

share|cite|improve this question
Isn't it just the identity map? You need it to be $k$-linear, so you need $\Lambda^0(f)(a) = \Lambda^0(f)(a\cdot 1) = a\Lambda^0(1)$, so it is completely determined by the image of $1$. And $\Lambda^k(f)(v) = \Lambda^k(f)(1\cdot v) = \Lambda^0(f)(1)\wedge \Lambda^k(f)(v)$, so $\Lambda^0(f)(1)=1$. But I could be doing something unwarranted here. – Arturo Magidin Mar 28 '11 at 4:41
You are quite correct, Arturo. The same thing happens with the tensor algebra, or the symmetric algebra... Also, the image of $1$ has to be $1$ for the map to be a morphism of algebras, so one can skip the computation :) – Mariano Suárez-Alvarez Mar 28 '11 at 4:46
@Arturo. I was unaware that $\Lambda^k(f)(a\cdot v) = \Lambda^0(f)(a)\wedge\Lambda^k(v)$. Does that derivation still work if $f:V\to W$ is the zero map? – NymSudo Mar 28 '11 at 4:58
@Nymsudo: I derived that from the graded structure: if $a\in\Lambda^k(V)$ and $b\in\Lambda^{\ell}(V)$, then $a\wedge b = (-1)^{k\ell}(a\wedge b)\in\Lambda^{k+\ell}$. The fact that your map has to respect the graded structure gives that equality. – Arturo Magidin Mar 28 '11 at 13:36
@Arturo: If you answer, I'll upvote to get this off the unanswered list. – Jonas Meyer Apr 2 '11 at 19:45

The map is the identity map. The map must be $k$-linear, so $$\Lambda^0(f)(a) = \Lambda^0(f)(a\cdot 1) = a\Lambda^0(1)$$ hence the map is completely determined by the image of $1$. But since it is a morphism of algebras, the map must send $1$ to $1$, so $\Lambda^0(1)=1$, hence $\Lambda^0(f)(a) = a$ for all $a\in k$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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