# How to prove that the wedge product is the determinant (Spivak's Claim)

In the book I am using now, Spivak's A comprehensive introduction to differential geometry volume 1 I have a question on page 205. Because he says the following

I think this is equivalent to say the following Let $\{e_i\}$ be the standard basis for $\mathbb{R}^{n}$ and let $\{e_i^{\ast}\}$ be dual basis then I want to conclude that $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}$ is the determinant tensor because I have seen that is books or courses this is given as the definition but for me in dimension 2 is little bit different,

$$v_1^{\ast} \wedge v_2^{\ast}=v_1^{\ast} \otimes v_2^{\ast}-v_2^{\ast} \otimes v_1^{\ast}$$

I know that somehow this could be interpret as the determinant but I can't figure out how can I prove the equivalence of the definitions and what Spivak is claiming in that part of the book.

I was trying to figure out this knowing how the wedge product $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}$ evaluate on the standard basis, and using skew symmetry and multi-linearity, but I'm still puzzled.

Attempts

• I was trying to do something of the following sort

We take $T \in \Lambda^{k}(V^{k})$ and $\{v_i\}$ a basis for V, then we can form k-tuples but we discard the ones that are $v_i\times ... \times v_i$ since this is zero so $T$ is uniquely determined by $T(v_1,...,v_k)=c$. But then

$$c(v_1^{\ast} \wedge v_2^{\ast} \wedge ... \wedge v_n^{\ast})$$

is in $\Lambda^{k}(V^{k})$ and $c(v_1^{\ast} \wedge v_2^{\ast} \wedge ... \wedge v_n^{\ast})(v_1,...,v_k)=c$ so that implies that $T=c(v_1^{\ast} \wedge v_2^{\ast} \wedge ... \wedge v_n^{\ast})$

But I don't know if this is right or how to perform this in my case.

• The reason because of I was trying to use the evaluation in the standard basis is because I have the following lemma

Lemma: Let $A: V \to V$ be a linear map. Then if $w_i=Av_i$, $v_1,v_2 \in V$ then

$$v_1^{*} \wedge v_2^{*}(w_1,w_2)=detA (v_1^{*} \wedge v_2^{*})(v_1,v_2)$$

• May be another way to prove this is to prove that that the wedge product satisfies the definition of determinant given in the book Second Year Calculus by Bressoud but by definition the wedge product is linear and the conditions that $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}(e_1,...,e_n)=1$ holds (I don't know how to prove that) the same for the condition that $e_1^{\ast} \wedge e_2^{\ast} \wedge ... \wedge e_n^{\ast}(e_1,e_2,...,e_2,e_n)=0$ So I don't know if this is a good way to proceed.
• This follows almost immediately from the fact that $\dim \Lambda^n (\Bbb R^n)^* = 1$. – Travis Apr 24 '16 at 19:05
• Ok, but Can you elaborate please? I don't get it :) – user162343 Apr 24 '16 at 19:05
• Perhaps you can be more specific about what you find unclear? – Travis Apr 24 '16 at 19:11
• Let me edit my post :), just a second, but Have you down voted my question? – user162343 Apr 24 '16 at 19:11
• @user162343: I wasn't the one who downvoted you. But something that I have noticed is that the community frowns upon requests for a "detailed" answer or proof, etc. It sounds like you are fishing for someone to do your homework. – Martin Argerami Apr 24 '16 at 19:16

As $\dim \Lambda^k (V^n) = \binom{n}{k}$, if $k = n$, then clearly $\dim \Lambda^n (V^n) = 1$. If $\lambda \in \mathcal L(V,V)$ (set of all linear transformations) then $\lambda^* : \Lambda^n (V^n) \to \Lambda^n (V^n)$ is a linear transformation between one-dimensional vector spaces, and so it must be a multiplication by an element in $\mathbb R$. Thus there is a unique number (call it) $\det (\lambda) \in \mathbb R$ such that $$\lambda^*(\omega) = \det (\lambda) \omega$$

where $\omega \in \Lambda^n(V^n)$.

Now if $A = (a^{i}_{j})$ is the matrix of $\lambda \in \mathcal L (V,V)$ and taking $\{e_1,\ldots, e_n\}$ some basis with dual $\{e^1,\ldots, e^n\}$ (I'm using $e^i$ instead $e_i^*$ just to avoid typing) then we have

\begin{align}\lambda^*(e^1 \wedge \ldots \wedge e^n) (e_1, \ldots, e_n) &= (e^1\wedge \ldots \wedge e^n)\lambda(e_1, \ldots, e_n)\\&= (e^1 \wedge \ldots \wedge e^n) (\lambda e_1, \ldots, \lambda e_n)\\&= \det (e^i (\lambda e_j)) = \det A\end{align}

since $\lambda (e_i) = \sum a^i_j e_j$. On the other hand, $$\lambda^* (e^1 \wedge \ldots \wedge e^n) = (\det \lambda )(e^1 \wedge \ldots \wedge e^n)$$

and as $(e^1 \wedge \ldots \wedge e^n) (e_1,\ldots, e_n) = 1$ we get that $\det (\lambda) = \det A$.

Edit: Here is a Lemma

Let $\alpha^1 , \ldots, \alpha^k$ elements of $\Lambda^1(V^n)$ and $v_1, \ldots, v_k$ vectors in $V$. Then $$(a^1 \wedge \ldots \wedge \alpha^k)(v_1, \ldots, v_k)=\det A$$

where $A = (a^i_j)$ is the $k \times k$ matrix whose $ij-$th entry is $a^i_j = \alpha^i (v_j)$.

Proof: (idea) Show that $$\alpha \wedge (\beta \wedge \gamma) = \frac{(k_1 + k_2 + k_3)!}{k_1!k_2!k_3!} \mathrm {Alt} \, (\alpha \otimes (\beta \otimes \gamma))$$

by inductive application you get $$a^1 \wedge \ldots \wedge \alpha^k = k! \mathrm {Alt} (\alpha^1 \otimes \ldots \otimes \alpha^k)$$

Thus $$a^1 \wedge \ldots \wedge \alpha^k(v_1, \ldots, v_k) = \sum_{\sigma} \mathrm{ sign} (\sigma)\,\,\alpha^1(v_{\sigma(1)}) \cdots \alpha^k(v_{\sigma(k)}) = \det A$$

where $\sigma \in S_k$ is a permutation.

Note: Here $\alpha \in \Lambda^{k_1}(V^n)$, $\beta \in \Lambda^{k_2}(V^n)$, $\gamma \in \Lambda^{k_3}(V^n)$.

• I don't get the last part of the computation for $\lambda^{*}$ :) – user162343 Apr 24 '16 at 20:42
• Good hahaha :) just give me some time to check it and if I have some question can I let you know ? – user162343 Apr 24 '16 at 20:43
• Sure, I hope that I may help you. – Aaron Maroja Apr 24 '16 at 20:45
• Right so just give me some time and I'll tell you right ? :) – user162343 Apr 24 '16 at 20:46
• Just a question your $v$ should be my standard basis right ? – user162343 Apr 24 '16 at 20:47