Understanding of graded algebra

Question

I am recently learning from Loring W. Tu's An Introduction to Manifolds the concept graded algebra, which is used for introducing exterior algebra. I don't understand the following definition:

An algebra $A$ over a field $K$ 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 $A^{k+l}$.

Here are the questions:

• What does $k$ in $A^k$ mean? Is it a superscript or the power of $A$? (If it is the power, what does $A^0$ mean?)
• Since I don't know much but some very basic knowledge in abstract algebra, I am trying to understand the concept with some simple examples. (I don't quite understand the wiki article of graded algebra). Is there any example as simple as the linear space ${\mathbb R}^n$ for the graded algebra? And what does it look like concretely?

Answer 1

1. It's a superscript.

2. The standard example is the polynomial ring $K[x_1, ... x_n]$, which is graded by total degree. That is, you can take $A^k$ to be the subspace of homogeneous polynomials of degree exactly $k$. In fact this is the free graded $K$-algebra on $n$ elements of degree $1$.

People who talk about graded algebras often don't bother to point out that specifying a grading is essentially the same thing as specifying a nice representation of the multiplicative group $K^{\times}$ on the algebra. An element $a \in K^{\times}$ acts on elements of degree $k$ by $x \mapsto a^k x$. The first axiom of a graded algebra says that the algebra splits up into a direct sum of irreducible representations, and the second axiom says that the action of $K^{\times}$ respects multiplication.

Answer 2

The most "ordinary" object that you have seen that is graded is the polynomial ring $\mathbb{F}[x]$ over a field, where the decomposition is into $\bigoplus_{i=0}^{\infty}x^i\mathbb{F}$. (It's the simplest case of Qiaochu Yuan's suggestion above).

I wanted to also add what a grading does for you: if you have two elements from summands of grades m and n, say $3x^n$ and $-5x^m$, then the grading tells you where the product lands: in the $x^{m+n}\mathbb{F}$ summand. That's a little better than ordinary rings, because it can be hard to predict where products land, sometimes!

The polynomial example looks a little humdrum, but it gets interesting when your grading is over something other than the natural numbers. Gradings can also be over the integers, a group, and even semigroups. Semigroup algebras are all naturally graded by their associated semigroups. (In fact, the polynomial ring is the semigroup algebra for the semigroup $\{1,x,x^2\dots\}$

The tensor algebra of a vector space is naturally graded by "type", and that is probably what you are encountering when reading about manifolds. I suppose $\mathbb{Z_2}$ graded algebras are also showing up there.

Answer 3

To add to Qiaochu's answer, often one can "explain" the existence of a grading on a mathematical object by giving a representation of the "circle group" (analytically, $S^1$, algebraically, the multiplicative group). The idea is that a representation $V$ of the circle group (here I'll use the analytic language) canonically decomposes into Fourier modes: the irreducible representations of $S^1$ are parametrized by the characters $\chi_k: t \mapsto t^k, S^1 \to \mathbb{C}^*$, and any nice representation $V$ splits as $$V = \bigoplus_{k \in \mathbb{Z}} V[k]$$ where $V[k]$ denotes the subspace of vectors $v \in V$ such that $S^1$ acts on $v$ via the one-dimensional character $\chi_k$. Conversely, given a graded vector space $ V = \bigoplus_{k \in \mathbb{Z}} V_k$, one defines a circle action by letting $S^1$ act on the $k$th piece by $\chi_k$.

Answer 4

In general $k$ is just an index, but often it is also some kind of power.

For example, for a vector space $V$ over a field $K$ one defines the tensor algebra $T(V)$ as $$T(V) = V^0 \oplus V^1 \oplus V^2 \oplus \cdots$$ where $$\begin{align} V^0&=K,\\ V^1&=V,\\ V^2&=V\otimes V,\\ V^3&=V\otimes V\otimes V, \\ \vdots \end{align}$$ and the multiplication map is $\otimes:V^m \times V^n \to V^{m+n}$ extended to the whole of $T(V)$ by linearity in both arguments.