# Why is the space of differential forms $\bigoplus_{p=0}^n \Lambda_x^p$?

In Wald's book "General Relativity", the space $$\Lambda_x$$ of differential forms at a point $$x$$ is worked out in the following manner: Let $$M$$ be an $$n$$-dimensional manifold. The vector space of all $$p$$-forms at a point $$x \in M$$ is given by $$\Lambda_x^p$$. The vector space $$\Lambda_x$$ of all differential forms (not limited to a specified degree $$p$$) is given by the direct sum:

$$\Lambda_x =\bigoplus_{p=0}^n \Lambda_x^p$$

However, from my understanding, the direct sum $$A \oplus B$$ of two spaces $$A$$ and $$B$$ consists of all ordered pairs $$(a,b)$$ where $$a \in A$$ and $$b \in B$$, with the additional structure:

$$(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)$$

Wouldn't this mean that an element from $$\omega \in \Lambda_x$$ would take the form:

$$\omega = (\omega_1, \omega_2, \dots , \omega_n)$$

where $$\omega_1$$ is a 1-form, $$\omega_2$$ a 2-form, and so on? This to me doesn't seem to be the space of all differential forms at $$x$$, unless all $$\omega_i$$ are 0 except one for each $$\omega$$. In fact, it seems to be a much larger space, since it can contain multiple differential forms of different degrees. My first guess was that a $$p$$-form and a $$q$$-form would be combined via the map $$\bigwedge: \Lambda^p_x \times \Lambda^q_x \mapsto \Lambda^{p+q}_x$$, but then we are no longer limited to the degree $$n$$. Why is this written as a direct sum rather than, for example, a union?

$$\Lambda_x = \bigcup_{p=0}^n \Lambda_x^p$$

• It might be useful to note where you learned the notation from, so that potential answers can work with potential nuance in the rest of the text Dec 30, 2021 at 11:42
• @CalvinKhor Thanks! I'm reading "General Relativity" by Wald, and this is taken from Appendix B. I don't know whether it will help, but I added it to the question for clarification.
– Max
Dec 30, 2021 at 11:47
• My suspicion is this is what's called the exterior algebra of the tangent space at the point $x$. Remember, when a space is finite dimensional of dimension say $n$ then any $p$ form with $p>n$ is necessarily 0. Dec 30, 2021 at 12:05

I can give a brief description of the exterior algebra if that helps. There are a few ways to go about this but I will take the one that is hopefully the most clear.

Let's start with a vector space $$V$$. Remember in our context $$V$$ is the tangent space to the manifold at a given point and we actually get a whole family of spaces that vary in a smooth way as you move around the manifold (change the base point of the tangent space). Now, we wish to consider all alternating forms on $$V$$ which are by definition multi-linear maps that satisfy an alternating condition. For example a 2-form satisfies $$\omega(x_1,x_2)=-\omega(x_2,x_1)$$ and an $$k$$-form satisfies $$\omega(x_{\sigma(1)},...,x_{\sigma(k)})=sgn(\sigma)\omega(x_1,...,x_k)$$ where $$\sigma$$ is a permutation on $$k$$ letters and $$sgn(\sigma)$$ (the signature) is either 1 or -1 depending upon $$\sigma$$.

Remark The alternating condition forces any $$m$$-form with $$m>n$$ variables to be exactly $$0$$. You can see this by realizing that if a set of vectors $$v_1,...,v_k$$ are linearly dependent, then for any $$k$$-form $$\eta$$ we have $$\eta(v_1,...,v_k)=0$$. You can show this by induction on a 2-form. For example $$\eta(x,cx)=c\eta(x,x)$$ now interchange the two entries and the alternating condition forces $$\eta(x,x)=-\eta(x,x)$$ which means $$\eta(x,x)=0$$. What do you know about a set of vectors $$v_1,...,v_m$$ in a vector space $$V$$ with $$dim(V)=n$$ and $$m>n$$?

Now we wish to construct an object with a multiplication that contains all such maps. We see that there will be $$n+1$$ distinct classes of objects: the $$k$$-forms for $$k=0,...,n$$.
We define the collection $$\Lambda^k(V)=\{\text{alternating}\quad k-\text{linear maps on}\quad V\}$$ You can check that $$\Lambda^k(V)$$ is a vector space and that its dimension is $$n\choose k$$.

It makes sense to organize them as $$n+1$$-tuples where the $$k$$'th entry is $$\Lambda^k(V)$$. The multiplication comes from the wedge product which you mentioned in your post. The wedge product sends you up the ladder of forms but necessarily stops at $$n$$ because of the above remark. So we can consider the object $$\Lambda(V)=\bigoplus_{k=0}^n \Lambda^k(V)$$ What does an element $$\lambda\in \Lambda(V)$$ look like? $$\lambda=(\omega_0,...,\omega_n)$$ with $$\omega_k\in \Lambda^k(V)$$.

In the context of differential geometry, we now have such a collection defined at every point on the manifold $$M$$ and the coefficients are not just scalars but can be smooth functions on $$M$$. To understand these objects a bit more formalism is required such as the idea of tangent bundles and so on but I won't get into this here.

Why are differential forms important? Well, they can be integrated over the manifold in a straightforward way and allows one to define volume on a manifold. Of course, in the context of physics they are important because they are tensors and tensors allow one to discover invariant object on $$M$$ (coordinate independent). You will see that Einstein's field equations of gravitation are statements about certain tensors on spacetime; the Ricci curvature tensor $$R_{\mu \nu}$$ and energy momentum tensor $$T_{\mu \nu}$$. These are not necessarily differential forms but they can be understood and constructed with these concepts.

• Thank you for your response! I read through your post quickly as I'm outside at the moment but will read it again thoroughly when I get home. Before I accept it (and if this question doesn't make sense, I'll get back to you when I've read it thoroughly), couldn't several forms be represented by the same object in $\Lambda(V)$? For example, $(0, \omega_1, \omega_2, 0)$ and $(0, 0, 0, \omega_3)$, if the combination of $\omega_1$ and $\omega_2$ maps into $\omega_3$? I think I understand the rest of your answer but am just a bit confused as to why this way of organizing it makes sense.
– Max
Dec 30, 2021 at 14:24
• @Max What combination of $\omega_1, \omega_2$ do you mean? A wedge product could make this happen but that takes you up to the fourth coordinate. If you write the objects with respect to a basis this can help. Dec 30, 2021 at 18:17
• Exactly, I referred to the case $\omega_3 = \omega_1 \wedge \omega_2$. Basically, my thinking goes like this: Each element from $\Lambda(V)$ represents a specific differential form in $V$. The only way this made sense to me is if (for example) $(0, \omega_1, 0)$ is used to represent $\omega_1$, and $(0, 0, \omega_2)$ to represent $\omega_2$. But then, what is $(0, \omega_1, \omega_2)$? If it is $\omega_1 \wedge \omega_2$, then wouldn't that mean that the same differential form (in some cases) can be represented by several vectors in $\Lambda(V)$?
– Max
Dec 30, 2021 at 20:13
• Thanks for all your help! I think it's clear now that I misinterpreted the space $\Lambda(V)$, and that the way to understand this further is to look into graded algebra (and tensor algebra). This does answer my initial question!
– Max
Dec 30, 2021 at 23:54
• By the way: math.stackexchange.com/a/2349194/203427 From this answer it seems like $\Lambda(V)$ consists of elements with mixed ordered forms (for example $\omega_1 + \omega_2$). For some reason I didn't consider this possible, but if this is true, then I understand why it is a direct sum (in this case, graded algebra).
– Max
Dec 31, 2021 at 0:45

I would consider it as the space of formal sums of differential forms. So an element in the space might be $$3dx_1 \wedge dx_2 \wedge dx_3 - 19dx_1 \wedge dx_2 + 4$$. The operations like $$\wedge$$ and addition still make sense if you extend them by linearity.