Exterior Algebra of smooth differential forms

Question

I'm a little bit confused about the exterior algebra of smooth differential forms $\Omega(M)$ on a manifold M. The definition of k-forms is clear to me, but I don't understand how to put them together, s.t. they form $\Omega(M)$ so to speak. Maybe you can help me to get rid of my problems:

First of all there is some confusion about the definition of $\Omega(M)$. Some people write $\Omega(M):=\bigoplus\limits_{k=0}^{\dim(M)}\Omega^k(M)$ and some $\Omega(M):=\sum\limits_{k=0}^{\dim(M)}\Omega^k(M)$. I never saw the second notation before, are they both the same by definition or is there a different meaning by the second one?

Furthermore, if we accept the definition $\Omega(M):=\bigoplus\limits_{k=0}^{\dim(M)}\Omega^k(M)$ then the elements of $\Omega(M)$ will consist of tuple like $(\omega_0,...,\omega_{\dim(M)})$, whereas $\omega_k\in\Omega^k(M)$. How do I extend the definition of the wedge product of single forms, i.e. $w_i\wedge\omega_j$, to elemts of the algebra, i.e. $(\omega_0,...,\omega_{\dim(M)})\wedge (\alpha_0,...,\alpha_{\dim(M)})=?$ I suggest that one writes the elements as "formal sums" like $(\omega_0,...,\omega_{\dim(M)})=:\omega_0+...+\omega_{dim(M)}$ and then extend the wedge product bilinearly. Is that right?

I hope someone can help me by answering my two questions.

Regards

Answer 1

Honestly I do not remember the notation $\sum$ you introduce above.

I will be quite sloppy with notation trying to keep all simple. Let $n$ denote the dimension of $M$, a finite dim. real manifold. The $k$-differential forms on $M$ are the elements of the space $\Omega^{k}(M)$; it follow from the very definition that such $k$-diff. forms can be locally written as $\omega=\omega_{i_1\dots i_k}(p) dx^{i_1}\wedge\dots\wedge dx^{i_p}$, where $p\in M$. $\omega_{i_1\dots i_k}(p)$ are smooth functions and $\{dx^{\bullet}\}$ denotes a basis of the finite dimensional vector space $\wedge^k T^{*}_p(M)$. From the very definition of wedge product it follows that $ dx^{i_k}\wedge dx^{i_l} =-dx^{i_l}\wedge dx^{i_k}$. So $\Omega^{k}(M)=0$ if $k>n$ (try to write a wedge product of $dx^{\bullet}$'s with more than $n$ terms: by antisymmetry you can prove that the product is zero). The direct sum $\Omega(M)=\oplus_{i=0}^n\Omega^{k}(M)$ is a graded algebra; the "grading" is the integer $k$ we introduced above. The algebra structure is given by the associative product $*:\Omega^{k}(M)\otimes\Omega^{l}(M)\rightarrow \Omega^{k+l}(M)$, $\omega_k *\omega_l:=\omega_k\wedge\omega_l$. Note that the product "respects" the grading: the structure $\Omega^{k}(M)\otimes\Omega^{l}(M)$ has degree $k+l$ (it follows from its very definition as tensor product), which is the same degree of $\Omega^{k+l}(M)$. The product is then extended bilinearly: $(\omega_1+\dots+\omega_k) *\omega_l:=\omega_1\wedge\omega_l+\dots+\omega_k\wedge\omega_l$ and $\omega_s*(\omega_1+\dots+\omega_t) :=\omega_s\wedge\omega_l+\dots+\omega_s\wedge\omega_t$.

Hope this can help

Answer 2

The answer is yes to both questions, though I don't really know why someone's using the $\sum$ notation -- where did you see it?