# The Unapologetic Mathematician

## Mathematics for the interested outsider

We’re about to talk about certain kinds of algebras that have the added structure of a “grading”. It’s not horribly important at the moment , but we might as well talk about it now so we don’t forget later.

Given a monoid $G$, a $G$-graded algebra is one that, as a vector space, we can write as a direct sum

$\displaystyle A=\bigoplus\limits_{g\in G}A_g$

so that the product of elements contained in two grades lands in the grade given by their product in the monoid. That is, we can write the algebra multiplication by

$\displaystyle\mu:A_g\otimes A_h\rightarrow A_{gh}$

for each pair of grades $g$ and $h$. As usual, we handle elements that are the sum of two elements with different grades by linearity.

By far the most common grading is by the natural numbers under addition, in which case we often just say “graded”. For example, the algebra of polynomials is graded, where the grading is given by the total degree. That is, if $A=R[X_1,\dots,X_k]$ is the algebra of polynomials in $k$ variables, then the $n$ grade consists of sums of products of $n$ of the variables at a time. This is a grading because the product of two such homogeneous polynomials is itself homogeneous, and the total degree of each term in the product is the sum of the degrees of the factors. For instance, the product of $xy+yz$ in grade $2$ and $x^3+xyz+yz^2$ in grade $3$ is

$\displaystyle (xy+yz)(x^3+xyz+yz^2)=x^4y+x^3yz+x^2y^2z+2xy^2z^2+y^2z^3$

in grade $5=2+3$.

Other common gradings include $\mathbb{Z}$-grading and $\mathbb{Z}_2$-grading. The latter algebras are often called “superalgebras”, related to their use in studying supersymmetry in physics. “Superalgebra” sounds a lot more big and impressive than “$\mathbb{Z}_2$-graded algebra”, and physicists like that sort of thing.

In the context of graded algebras we also have graded modules. A $G$-graded module $M$ over the $G$-graded algebra $A$ can also be written down as a direct sum

$\displaystyle M=\bigoplus\limits_{g\in G}M_g$

But now it’s the action of $A$ on $M$ that involves the grading:

$\displaystyle\alpha:A_g\otimes M_h\rightarrow M_{gh}$

We can even talk about grading in the absence of a multiplicative structure, like a graded vector space. Now we don’t even really need the grades to form a monoid. Indeed, for any index set $I$ we might have the graded vector space

$\displaystyle V=\bigoplus\limits_{i\in I}V_i$

This doesn’t seem to be very useful, but it can serve to recognize natural direct summands in a vector space and keep track of them. For instance, we may want to consider a linear map $T$ between graded vector spaces $V$ and $W$ that only acts on one grade of $V$ and with an image contained in only one grade of $W$:

\displaystyle\begin{aligned}T(V_i)&\subseteq W_j\\T(V_k)&=0\qquad k\neq i\end{aligned}

We’ll say that such a map is graded $(i,j)$. Any linear map from $V$ to $W$ can be decomposed uniquely into such graded components

$\displaystyle\hom(V,W)=\bigoplus\limits_{(i,j)\in I\otimes J}\hom(V_i,W_j)$

giving a grading on the space of linear maps.

October 23, 2009 - Posted by | Algebra, Linear Algebra, Ring theory

1. To be fair, superalgebras and $\mathbb{Z}/2\mathbb{Z}$-graded algebras aren’t the same thing; I’m told the tensor product structure on the former is different.

Comment by Qiaochu Yuan | October 23, 2009 | Reply

2. Do you have a source to cite? I’ve never heard anything like that, and in fact common sources like Wikipedia start off by saying that the two are the same thing.

Comment by John Armstrong | October 23, 2009 | Reply

3. Most people who speak of (Z mod 2)-graded algebras actually intend the same thing as superalgebras. It’s all in the symmetry of course. If G is a commutative monoid, then there’s a symmetry isomorphism on G-graded abelian groups (or G-graded vector spaces) which just swaps the tensor factors of homogeneous components and does nothing else, and when specialized to G = Z mod 2, the monoids therein could be called (Z mod 2)-algebras as Qiaochu is using the term. But there’s another (much more commonly used) symmetry where a factor of (-1)^{pq} is introduced, where p and q are the homogeneous degrees. Monoids in that symmetric monoidal category are what Qiaochu is calling superalgebras.

(Naturally, the concept of monoid doesn’t involve the symmetry. However, many algebra constructions will involve the symmetry, and there the distinction is important.)

Comment by Todd Trimble | October 23, 2009 | Reply

4. Second Z mod 2 in fourth line: should be “(Z mod 2)-graded”.

Comment by Todd Trimble | October 23, 2009 | Reply

5. What Todd said, with the addition that the whole issue here is whether you ‘just’ want algebras that decompose into two parts, or whether you want the parts to commute/anti-commute dependent on degree.

In the latter case – which pops up all over the place – you really want the signed Z/2-action instead of the trivial one on your symmetric monoidal categories.

Comment by Mikael Vejdemo Johansson | October 23, 2009 | Reply

6. I haven’t remotely talked about the symmetries yet. At the level of just “graded algebras”, they’re the same thing. Complaining that I don’t talk about graded commutativity (of one particular sort) when I introduce graded algebras is like complaining I don’t talk about commutativity when I introduce algebras.

Comment by John Armstrong | October 23, 2009 | Reply

7. You’re right: no, you haven’t, and yes, they are. The distinction between just Z/2-graded and “super” that Qiaochu is referring to is still worth observing though.

For example, the statement that the algebra of differential forms (which you are coming to soon?) is supercommutative would no doubt be confusing unless the distinction were observed. So it’s good to talk about this in comments.

Comment by Todd Trimble | October 23, 2009 | Reply

8. Well, exterior algebras are natural-graded (and graded-commutative, within that context). Yes, it’s possible to “roll” a natural-graded algebra into a $\mathbb{Z}_2$-graded one, but it loses information and isn’t the natural decomposition to work with here anyway.

Comment by John Armstrong | October 23, 2009 | Reply

9. On a different note, the grading on $A=R[x_1,\ldots,x_n]$ in the group $\mathbb{Z}_{\geq 0}^n$ given by $x^\alpha\in A_\alpha$ is very useful in computational algebraic geometry, and choosing a total ordering on $\mathbb{Z}_{\geq 0}^n$ is what’s needed to make division by an ideal work out, and many other important computations.

Comment by Charles Siegel | October 23, 2009 | Reply

10. Oh, sure. If and when I ever get around to algebraic geometry, I’ll be talking about that grading.

Comment by John Armstrong | October 23, 2009 | Reply

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13. The piece at the end is incorrect, because an element of a direct sum must be a \emph{finite} sum of its components; as it is stated the identity map on a vector space graded of the natural numbers (or any infinite set) does not have a decomposition into graded components. Replacing the direct sum by a direct product will not do either, because then most collection of components would fail to combine to a linear map on the graded space at all. What one seems to need is that the image of every single graded component lives in finitely many graded components, but images of different graded components are unrelated.

“For instance, we may want to consider a linear map T between graded vector spaces V and W that only acts on one grade of V and with an image contained in only one grade of W:

We’ll say that such a map is graded (i,j). Any linear map from V to W can be decomposed uniquely into such graded components

\displaystyle\hom(V,W)=\bigoplus\limits_{(i,j)\in I\otimes J}\hom(V_i,W_j)

giving a grading on the space of linear maps.”

Comment by Marc van Leeuwen | December 4, 2009 | Reply

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