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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?

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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
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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

1 Answer 1

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 $a\in k$.

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