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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:

  1. Sending $p \in M$ to a vector $v(p) \in V$.
  2. 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?

CBBAM
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  • What is your definition of a differential $k$-form with values in $V$? What you describe is not consistent with the usual definition (which is Tu's definition). – Michael Albanese Nov 13 '23 at 22:14
  • @MichaelAlbanese I view it as a map that sends $p \in M$ to another map $\omega_p: T_pM \times \cdots \times T_pM \rightarrow V$. where the domain of $\omega$ has $k$ copies of $T_pM$. – CBBAM Nov 13 '23 at 22:25
  • I didn't realise this the first time around. Did you mean for $V$ to be a vector space or a vector bundle? Tu's definition is for $V$ a vector bundle. – Michael Albanese Nov 13 '23 at 22:29
  • @MichaelAlbanese I am learning this for applications to vector bundles, but I am taking $V$ to just be a vector space for now to keep it simple. – CBBAM Nov 13 '23 at 22:34
  • You need to learn tensor products in a setting other than the tensor product of multilinear maps, yes. – Ted Shifrin Nov 13 '23 at 23:29
  • @TedShifrin I will read more on tensor products in general. Is it necessary to define vector-valued forms using tensor products? – CBBAM Nov 14 '23 at 00:08
  • Not if you’re comfortable writing expressions like $\omega,v$, but where does this animal actually live? – Ted Shifrin Nov 14 '23 at 01:39
  • @TedShifrin I see, so it is more to be specific about what object you're working with? – CBBAM Nov 14 '23 at 02:17
  • It is to define mathematically/rigorously the space these things live in, yes. And then there are standard operations and constructions that are mathematically defined. – Ted Shifrin Nov 14 '23 at 02:21
  • @TedShifrin I think I need to read more and then come back to this question because right now I don't see how the tensor product makes something like $\omega v$ any less rigorous than if a tensor product was used. – CBBAM Nov 14 '23 at 04:15
  • The simplest vector valued two form is the tensor product of a vector and a two form: $\partial_x\otimes\omega,.$ This can also be seen as a $(1,2)$ tensor and is therefore nothing really new. MTW devoted a whole section 14.5 to this. I don't quite see how else a vector valued two form could be defined. – Kurt G. Nov 14 '23 at 08:21
  • @KurtG. What is confusing me is why do we have to use a tensor product to write $\partial_x \otimes \omega$ and note simple $\partial_x \omega$? I get the sense that the latter is somehow not properly defined but that is what I am having trouble seeing. Also, what is MTW? – CBBAM Nov 14 '23 at 16:37
  • Yes indeed. If you don't like the orthodox tensor product definition you may want to properly define another product between $\partial_x$ and $\omega,.$ MTW is the "phone book" :) known as Misner, Thorne & Wheeler, Gravitation. As soon as I understood that a vector valued two form is simply a good old $(1,2)$ tensor I was happy and pressed on. – Kurt G. Nov 14 '23 at 17:25
  • One buzz word from algebra is using tensor product (in the setting of ring theory) to do "extension of scalars." This is the sense in which I view tensor product for vector bundle (or space)-valued forms, for example. – Ted Shifrin Nov 14 '23 at 18:12
  • @KurtG. Thank you, I will take a look at that reference. – CBBAM Nov 14 '23 at 18:29
  • @TedShifrin I see now, so instead of scalar multiplication we have an "extended" version which informally says the first term in the tensor product does one thing and the second does another but they're still one object, and so in this way we are sort of extending scalar multiplication. Is this the "extension of scalars" idea you were describing? – CBBAM Nov 14 '23 at 18:30
  • For example, if you have a $\Bbb Z$-module $M$, then $M\otimes_{\Bbb Z}\Bbb Q$ turns it into a $\Bbb Q$-vector space. So we're extending the scalars from integers to rational numbers. Now we're extending from $\Bbb R$-valued differential forms to $V$-valued differential forms (since $V$ is an $\Bbb R$-module). – Ted Shifrin Nov 14 '23 at 18:37
  • @TedShifrin Ah I see that makes a lot of sense now! Thank you. So one of the roles of the tensor product is (at least in some sense) to swap or generalize the scalar coefficients in vector fields? – CBBAM Nov 14 '23 at 19:08
  • In tensors of all types, in fact, yes. With differential forms you do have the exterior derivative for scalar-valued differential forms, and you then wonder what to do in general. With values in a fixed vector space, you can define $d$ using a basis and check the answer is independent of choice, but with values in a vector bundle you'll need a connection on the bundle. – Ted Shifrin Nov 14 '23 at 19:10
  • @TedShifrin So to summarize everything I know about tensor products, they have two uses (1) to generalize the scalars in a tensor field and (2) to be able to take "products" of tensors. Is there anything else they are used for? – CBBAM Nov 14 '23 at 19:48

1 Answers1

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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’:

  • an alternating multilinear map $\underbrace{V\times\cdots\times V}_{\text{$k$ times}}\to E$ (a common notation for the space of such maps is $\mathcal{A}^k(V,E)$).
  • a linear map $\bigwedge^k(V)\to E$, i.e an element of the vector space $\text{Hom}\left(\bigwedge^k(V),E\right)$
  • an element of $\bigg[\bigwedge^k(V)\bigg]^*\otimes E$, which is simply written as $\bigwedge^k(V)^*\otimes E$
  • an element of $\bigg[ \bigwedge^k(V^*)\bigg]\otimes E$, which is simply written as $\bigwedge^k(V^*)\otimes E$

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:

  • a smooth vector bundle morphism $\Phi:E\to F$ (i.e a smooth map $\Phi:E\to F$ such that for each $x\in M$, $\Phi$ restricts to a linear map of the fiber $E_x$ into the fiber $F_x$).
  • a smooth section $\tilde{\Phi}$ of the vector bundle $\text{Hom}(E,F)$ over $M$.
  • a smooth section $\tilde{\tilde{\Phi}}$ of the vector bundle $E^*\otimes F$ over $M$

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:

  • a smooth fiberwise-alternating multilinear morphism $(TM)^{\oplus k}\to E$
  • a smooth vector bundle morphism $\bigwedge^k(TM)\to E$
  • a smooth section of the vector bundle $\text{Hom}\left(\bigwedge^k(TM), E\right)$
  • a smooth section of the vector bundle, $\left[\bigwedge^k(TM)\right]^*\otimes E$
  • a smooth section of the vector bundle $\left[\bigwedge^k(T^*M)\right]\otimes E$

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 $\Omega^k(M;\Bbb{R})$.

peek-a-boo
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    anyway, the tldr of this answer is that much of the linear algebra really is ‘basic’ stuff, so you should review it from a textbook proper. – peek-a-boo Nov 15 '23 at 02:58
  • Thank you for the elaborate answer! What do you mean by ‘enrich the target space’? I am familiar with tensor products in the sense that for a vector spaces $V$ and $W$, we may "multiply" them using the tensor product $V \otimes W$ whose elements generally look of the form $v_i \otimes w_j$. I am having trouble connecting this picture with what you have in your answer. For example, when you say $\text{Hom}(W,E)\cong W^\otimes E$, why not write an element of the latter as simply $w^ e$? Since $w^* \in W^*$ is real valued and $e \in E$ is a vector this product makes sense. – CBBAM Nov 15 '23 at 08:00
  • I guess another way to phrase my confusion is that in this context I don't see what extra information tensor products give us that scalar multiplication already doesn't. – CBBAM Nov 15 '23 at 08:01
  • Regarding ‘enrich the target’, that’s just vague terminology; ignore if it makes you uncomfortable. Next, for each $\omega\in W^$ and each $e\in E$, we can indeed define an element of $T_{\omega,e}\text{Hom}(W,E)$ by setting $T_{\omega,e}(x):=\omega(x)\cdot e$, where the $\cdot$ is now scalar multiplication in $E$; and sure you would be justified to denote this linear map as simply $\omega(\cdot)e$ or just $\omega e$. BUT*, this is NOT the most general element of $\text{Hom}(W,E)$. – peek-a-boo Nov 15 '23 at 08:37
  • such a construction only gives you a very special class of linear maps (notice the range of such a linear map is a 1-dimensional subspace of $E$ (assuming $\omega\neq 0$), so if $\dim E>1,\dim W>1$ these definitely don’t describe all linear maps $W\to E$). You would need a linear combination of such maps. – peek-a-boo Nov 15 '23 at 08:40
  • so far we’ve only talked about ‘combining’ dual spaces with another vector space. What if I wanted to ‘combine’ two completely separate vector spaces $X,Y$? There is no ‘scalar multiplication’ between $X,Y$. Hence the need for tensor products (ok one thing you could do in finite-dimensional situations is identify $X\cong X^{*}=(X^)^*$ and repeat the above game; this is fine for vector spaces, but in more general algebraic situations, especially with modules over rings, where duals aren’t as nice and we don’t have bases generally, tensor products need to be dealt with separately). – peek-a-boo Nov 15 '23 at 08:43
  • So if we instead define $T \in \text{Hom}(W,E)$ as $T(x) = \sum_i \omega_i(x) \cdot e_i$ where the sum runs over a basis for $E$ would this now fix the problem of describing all linear maps and not just 1-dimensional ones? But I think I see your point now after reading your last comment. We're identifying $\omega \otimes e$ with $\omega \cdot e$, but this identification cannot be done for general algebraic objects and this is why the tensor product is needed. In this sense the tensor product really is a generalization of scalar multiplication. Have I gotten this right? – CBBAM Nov 15 '23 at 09:03
  • yes, that’s right (also, extension of scalars, as mentioned by Ted Shifrin above is a very good and concrete example to keep in mind for the power of tensor products). – peek-a-boo Nov 15 '23 at 09:40
  • Great, thank you for all your help! – CBBAM Nov 15 '23 at 09:53