I am reading Tu's book "An introduction to Manifolds". Specifically, I am reading about the the exterior algebra of multicovectors (subchapter 1.3) and I happened to come across a definition I didn't quite understand.
The author states that an algebra $A$ is said to be graded if it can be written as a direct sum $A=\bigoplus_{k=0}^{\infty} A^k$ of vector spaces over $K$, such that the multiplication map sends $A^k\times A^l$ to $A^{k+l}$. The notation $A=\bigoplus_{k=0}^{\infty} A^k$ means that each nonzero element of $A$ is uniquely a finite sum $$a=a_{i_1}+...+a_{i_m}$$ where $a_{i_j}\ne0\in A^{i_j}$.
First of all, I am not 100% certain I understand what a graded algebra is, so if anyone could elaborate a bit, it would be fantastic, and second, I am having difficult time in understanding why there is a need for having an index $j$ as a subscript of the index $i$ and therefore to understand the meaning of the equation written above.
Think of differential forms. For example, while $${\rm d}x + x^2{\rm d}x\wedge {\rm d}z - \cos(yz) \,{\rm d}x\wedge {\rm d}y\wedge {\rm d}z$$is not a differential form on its own right, it is a legitimate element of the exterior algebra of $\Bbb R^3$. But it equals the sum of three terms, each of which can be assigned a degree. And if you take the wedge product between a $k$-form and an $\ell$-form, you get a $(k+\ell)$-form, right?
He wants to do this abstractly.
The point is that $\bigoplus$ is not the same thing as $\bigcup$, so an element of $\bigoplus_{k\geq 0}A^k$ does necessarily belong to a single $A^k$, but instead equals a finite sum of factors (this is just the definition of direct sum, forget graded algebras and think of linear algebra), each of which can be assigned a degree (i.e., an element of $A^k$ is said to have degree $k$).
About the indices: the key thing is that by using $i_j$'s, he can keep only the nonzero terms and avoid writing superfluous zeros. Let's say that we have an element $a \in A$ which has no terms of degree greater than, say, $6$. He'd rather write $$a = a_1 + a_3 + a_6,\qquad a_1\in A^1\smallsetminus \{0\},\quad a_2\in A^3\smallsetminus\{0\},\quad a_6\in A^6\smallsetminus\{0\},$$with $m=3$, $i_1 = 1$, $i_2 = 3$, and $i_3 = 6$, than saying that $$a = a_1+a_2+a_3+a_4+a_5+a_6,\qquad a_i \in A^i\mbox{ for }i=1,\ldots,6,$$and having to say "but BTW, $a_2=a_4=a_5 = 0$".