For covariant tensors, why is it $\bigwedge^k(V)$, not $\bigwedge^k(V^*)$?

Question

In learning the very basics of differential geometry, I have seen the exterior product defined a couple of ways: First, I have seen it as the image of the covariant tensors (which I believe are essentially the $k$-fold tensor product of $V^*$ with itself) transformed by the mapping "Alt". Second, I have seen it defined somewhat more abstractly as the $k$-fold tensor product of $V$ with itself, modulo certain relations which make it anti-symmetric. (However, at this early stage it seems we have no need for contravariant tensors.) In either case, we write it $\bigwedge^k(V)$. Since we are focusing on covariant tensors, why do we not write $\bigwedge^k(V^*)$? Is there something I am missing which makes these two equivalent?

Answer 1

You're right: this is a notation failure (on the part of many differential geometry texts). The correct notation should be $\bigwedge^k (V^*)$, as you say. However, confusingly many texts write $\bigwedge^k(V)$ to mean $\bigwedge^k(V^*)$.

(edit by EA, 5:54. fixed a typo)

Answer 2

The reason why there should be a dual there is this: A covariant $r$ - tensor is by definition a multilinear map from $r$ copies $V \times \ldots \ldots \times V$ to the real numbers. We usually identify $$\operatorname{Mult}(V \times \ldots \times V,\Bbb{R}) \cong V^\ast \otimes \ldots \otimes V^\ast$$

via the map that sends an elementary tensor $\varphi_1 \otimes \ldots \otimes \varphi_r$ on the R.H.S. to the multilinear map defined by

$$\begin{eqnarray*} \varphi_1 \otimes \ldots \otimes \varphi_r : &V \times \ldots \times V& \to \Bbb{R} \\ &(v_1,\ldots,v_r)& \mapsto \varphi_1(v)\ldots \varphi_r(v_r)\end{eqnarray*}$$

The last expression on the right is multiplication of real numbers. Thus in a similar way, we identify

$$\operatorname{Alt}(V \times \ldots \times V,\Bbb{R}) \cong V^\ast \wedge \ldots \wedge V^\ast.$$

Remember: A covariant tensor has to eat some bunch of vectors and spit out a real number, so any sensible definition involving these should involve the dual!