Let $M$ be a manifold and $V$ a vector space. Then $\omega \in \Omega^k(M, V)$ is a differential $k$-form that takes values in $V$. I think about such forms as having two parts:
- Sending $p \in M$ to a vector $v(p) \in V$.
- Sending $p$ to $\omega_p$ which is an alternating tensor on $T_pM$.
Thus the image of such a differential form can be written as $\omega_p(X_1, \ldots, X_n) v$. Once evaluated, $\omega_p(X_1, \ldots, X_n)$ is a scalar and so $\omega_p(X_1, \ldots, X_n) v$ is well defined.
However in many sources, such as Tu's Differential Geometry, you see a much more complicated expression: $$\Omega^k(M, V) \equiv \Gamma \Bigg(\Big(\bigwedge^kT^*M\Big) \otimes V\Bigg). \tag{1}$$ I am confused on the use of the tensor product here, both understanding the above statement and also what advantages it has. The use of the tensor product in this definition seems very common and so I would like to learn why its necessary to define vector-valued forms using them.
EDIT: I should add that I am familiar with tensor products and usually think of this definition when I see them: $$(f \otimes g)(v_1, \ldots, v_{n+m}) = f(v_1,\ldots,v_n)g(v_{n+1}, \ldots,v_m).$$ Is this how the tensor products in (1) should be interpreted?
I have several answers which illustrate the ideas in many special cases (but once you ‘get’ the idea, you can do almost anything you want). Besides, you should indeed follow a textbook in detail rather than doing small pieces here and there because otherwise you’ll always be confused when a new situation pops up.
I’m sure there are more, but I’m too lazy to find them all. Anyway, here’s a consolidation of the ‘basics’.
1. Vector Space level.
The most important thing to understand is ALWAYS at the vector space level. Let $V,E$ be a finite-dimensional (real) vector spaces. Then, basic (multi)linear algebra tells us that the following are equivalent in terms of ‘information’:
The meaning of the $\otimes$ is that of a general tensor product of vector spaces: for any two vector spaces $X,Y$ one can construct a new vector space $X\otimes Y$ which satisfies a certain universal property (roughly, it amounts to taking ‘formal linear combinations’ of the symbols $x\otimes y$ for $x\in X,y\in Y$). This is covered in any good algebra textbook, so I won’t repeat it all here.
The equivalence of the first two is by the universal property of exterior powers (i.e almost by definition). The equivalence of the second and third is the usual Hom/tensor-product isomorphism that for finite-dimensional vector spaces $W,E$, $\text{Hom}(W,E)\cong W^*\otimes E$ (the way I think of this is that $W^*=\text{Hom}(W,\Bbb{R})$ consists of real-valued linear maps, so if we want $E$-valued linear maps, we need to ‘enrich the target space’, and this is done by tensor products). The equivalence of the third and fourth descriptions is by basic algebra involving duals and how it interacts with exterior powers. I’ll let you google these facts. Also, all these isomorphisms are canonical, so algebraically speaking, no one description is inherently better than any other.
Pedagogically, the first requires the least amount of preliminary vocabulary, so in that sense, it is the ‘simplest’. However, ‘easy to present’ doesn’t always mean ‘easy to work with’ or ‘quick to work with’. Once you gain familiarity with tensor products and exterior powers, you’ll see that you can do various things rapidly (tensor products of two maps $f_1:V_1\to E_1$ and $f_2:V_2\to E_2$ to get $f_1\otimes f_2:V_1\otimes V_2\to E_1\otimes E_2$, and similar stuff for exterior powers), so the other descriptions are also good to know. So, the fact that at the vector space level we have (at least) 4 ways of describing the same thing, it is no surprise that at the vector-bundle level, you get more ways/notations all of which are trying to say the same thing.
2. Vector Bundle stuff
Let $E,F$ be smooth vector bundles over $M$. The following objects convey the same information:
The equivalence of the first and second is simply by setting $\tilde{\Phi}(x)=\Phi|_{E_x\to F_x}$. Proving that smoothness of $\Phi$ is equivalent to smoothness of $\tilde{\Phi}$ is done writing things out in local vector-bundle charts. The equivalence of the second and third is by the linear-algebra analogue (done fiberwise). Symbolically, you could write this as \begin{align} \text{Mor}(E,F)\cong \Gamma\left(\text{Hom}(E,F)\right)\cong\Gamma(E^*\otimes F). \end{align}
3. Putting things together
Let $E$ be a vector bundle over $M$ and $k\geq 0$ an integer. In view of the vector-space discussion and and above vector bundle discussion, we can now define a exterior differential $k$-form on $M$ with values in $E$ to be any of the following objects:
Regardless of which ‘concrete’ implementation you decide to take, the space of all such $E$-valued $k$-forms on $M$ is often denoted simply $\Omega^k(M;E)$, and when $E=M\times V$ is a trivial vector bundle (with $V$ the common vector space), then we don’t write $\Omega^k(M; M\times V)$ but rather simply $\Omega^k(M;V)$. Finally, when $V=\Bbb{R}$ (i.e the usual (scalar) differential $k$-forms on $M$) we write $\Omega^k(M)$ rather than $\Omega^k(M;\Bbb{R})$.