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

Question

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?

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

Answer 1

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.

Answer 2

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.