$\mathbb{G}_m$ action on $\operatorname{Spec}A$ is equivalent to grading... what does coassociativity do?

Question

I can interpret the other axioms for a grading via the definition of a group scheme action appropriately (I think):

Let $H = k[x,x^{-1}]$. We are given a coaction $ \rho : A \to A \otimes H$, which is a morphism of $k$ algebras.

1. The counit axiom implies that the coaction map $\rho: A \to A \otimes H$ is injective (it explicitly gives an inverse), hence the direct sum structure on $A \otimes H$ given by $H = \bigoplus_{ n \in \mathbb{Z}} k_{x^n} \otimes_k A$ pulls back via $\rho^{-1}$ to give a direct sum decomposition of $A$: $A = \bigoplus_{n \in \mathbb{Z}} \rho^{-1} (k_{x^n} \otimes A)$.
2. That the coaction map is an algebra homomorphism implies that $A_n A_m \subseteq A_{n + m}$.

These appear to be all of the conditions of a $\mathbb{Z}$ grading on the ring $A$ - namely, that $A$ can be written as a direct sum of submodules graded by $\mathbb{Z}$ which multiply in the correct way. But I haven't used the coassociativity. So I suspect that I am being stupid in some way I would appreciate being pointed out.

Or perhaps the point of the coassiativity is to force the coaction to look like $f \to x^n \otimes f$, for $f$ a degree $n$ homogeneous element in the grading induced by $1$ (i.e. living in $\rho^{-1} ( k_{x^n} \otimes_k A)$? However, this seems to come from the conunit axiom anyway, since the counit evaluates $x^n$ to be $1$. ($f$ must be sent to some element in $k_{x^n} \otimes_k A$, i.e. some $ x^n \otimes_k g$, but conuit implies that it must be the case that $g = f$.)

As for the converse - starting with a graded ring, then defining a coaction by sending a degree $n$ homogeneous element to $x^n \otimes f$... it seems that the coassociativity more or less comes for free. So even if one had just a not necessarily associative action of $G_m$ on $\operatorname{Spec}A$ (i.e., a $k$-algebra coaction by $H$ satisfying just the couit axiom), it appears that it becomes automatically associative (it induces a grading, which then induces an associative coaction, which agrees at least on homogeneous elements). But this feels weird...

I'm really confused!

Answer 1

A bit late, but I worked this out just now.

The coaction is not just $f\mapsto x^n\otimes f$ (for some $n$ depending on $f$, or otherwise)! This only applies for homogeneous $f$. (Well, homogeneous according to the very grading we are trying to work out.) In general, $\rho(f)$ is a finite sum $\sum_i x^i\otimes f_i$, which I will write as $\sum_i f_ix^i$ from now on.

The counit axiom says that, for each element $f\in A$, $\rho(f)$, which is a finite sum $\sum_i f_i x^i$ in $A\otimes H$, satisfies $\sum_i f_i=f$ (thanks Daniel).

The coassociativity axiom applied to the same $f$ will give that $\rho(f_i)=f_ix^i$. This is a nontrivial statement because in general, all we know is that $\rho(f_i)$ is some sum $\sum_j (f_i)_j x^j$. But coassociativity says precisely that $\sum_i f_i(x^i\otimes x^i)=\sum_{i,j}(f_i)_j (x^j\otimes x^i)$, and matching terms, this says that $(f_i)_j=0$ for $i\neq j$, and $(f_i)_i=f_i$. So coassociativity is what guarantees that the pieces $f_i$ corresponding to some $f\in A$ are actually homogeneous. (Consistency of the grading, in some sense, if you like.)

Answer 2

Fix a base field $k$. Let’s prove carefully that if $\operatorname{Spec} A$ has a $\mathbb{G}_m = \operatorname{Spec} k[x,x^{-1}]$-action, then there is an associated grading, and conversely given a grading there is an associated $\mathbb{G}_m$-action. We will also prove these associations are mutually inverse. Hopefully this clarifies where coassociativity is used. (Namely, it is only required in defining the grading.)

A $k$-algebra $A$ is called graded, if there is a vector space decomposition $$A = \bigoplus_{m \in \mathbb{Z}} A_m$$ so that $A_m \cdot A_n \subseteq A_{m + n}$ for all $n \in \mathbb{Z}$.

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If $A$ is a $k$ algebra with a $\mathbb{G}_m$ action induced by $k$-algebra homomorphism $\rho: A \to A[x,x^{-1}]$ we will define the associated grading by setting $$A_m = \rho^{-1}(Ax^m).$$ To check this is a grading, we will first show that $$\sum_{m \in \mathbb{Z}} A_m = \bigoplus_{m \in \mathbb{Z}} A_m$$ and then that $\sum_{m \in \mathbb{Z}} A_m = A$ to conclude that $A$ admits this direct sum decomposition. Once this is done, we clearly have a grading because $$A_m \cdot A_n = \rho^{-1}(Ax^m) \cdot \rho^{-1}(Ax^n) \subseteq \rho^{-1}(Ax^m \cdot Ax^n) = \rho^{-1}(Ax^{n+ m}) = A_{n + m}$$ by the definition of a ring homomorphism.

To check that the sum is direct, we let $f \in A_n \cap A_m$ for some $n \neq m$. Then, $\rho(f) \in Ax^m \cap Ax^n = 0$, so $f \in \ker \rho$. However, as noted in the question, the counit axiom implies that $\rho$ is injective, so that in fact $f = 0$. Hence, the sum is direct

To check that $\sum_{m \in \mathbb{Z}}A_m = A$, we write $\rho(f) = \sum f_m x^m$ and evaluate along the counit ($x \mapsto 1$) to see that $f = \sum f_m$. To conclude, we must show that each of these $f_m$ is in $A_m$. This step is what requires coassociativity, as noted in the previous answer, which says exactly that $\rho(f_m) \otimes x_m = f_m x^m \otimes x^m$, so that in particular $\rho(f_m) = f_mx^m$. This implies that $f_m \in A_m$, as required.

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Conversely, if we’re given a grading $\bigoplus_{m \in \mathbb{Z}} A_m$ of a $k$-algebra $A$ we will define $\rho: A \to A[x, x^{-1}]$ by sending elements $f_m$ of $A_m$ to $f_m x^m$ and extending linearly. That is, $f = \sum f_m \in A$, it is sent to $\rho(f) = \sum f_m x^m \in A[x, x^{-1}]$. This linear map in fact defines a ring homomorphism because of the condition that $A_m \cdot A_n \subseteq A_{m + n}$.

Since the counit of $k[x, x^{-1}]$ sends $x \mapsto 1$, it is clear from the equation to start $f = \sum f_m$, that the composition $f \mapsto \sum_m f_m x^m \mapsto \sum f_m$ is indeed the identity, which confirms the unit condition of the action.

Similarly, since the comultiplication map of $k[x, x^{-1}]$ is $x \mapsto x \otimes x$, the coassociativity condition of the action is saying that the composite map $$f \mapsto \rho(f) = \sum f_m x^m \mapsto \sum \rho(f_m) \otimes x^m$$ is the same as $$f \mapsto \rho(f) = \sum f_m x^m \mapsto \sum f_m x^m \otimes x^m.$$ This then follows from the fact that $\rho(f_m) = f_m x^m$ by the definition of this homomorphism.

Hence $\rho$ defines an algebraic group action of $\mathbb{G}_m$ on $\operatorname{Spec} A$, by the definition found here, for example.

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Lastly, we check that these are mutually inverse associations. Given a grading $A = \bigoplus A_m$, we associate the action given by $\rho(f) = \sum f_m x^m$. To this action, we then associate a grading $A = \bigoplus B_m$ with $B_m = \rho^{-1}(Ax^m)$. We claim that $B_n = A_n$ for all $n$. Clearly, $A_n \subseteq B_n$. Conversely, let $f \in B_n$ and write $f = \sum f_m$ in the original grading of $A$. Then, $\rho(f) = f_n x^n$ because all the other coefficients are zero by the definition of $B_n$. As such, $f - f_n \in \ker \rho = 0$. So $f = f_n \in A_n$ as required.

Conversely, given an coaction $\rho: A \to A[x, x^{-1}]$ we associate a grading given by $A_m = \rho^{-1}(Ax^m)$. To this, we then associate the coaction $\tau: A \to A[x, x^{-1}]$ which sends $f = \sum f_m$ to $\tau(f) = \sum f_m x^m$. But now, by the definition of the grading and the counit axiom, for $f_m \in A_m$, $\rho(f_m) = f_mx^m$ for all $m$. Hence, $\rho$ and $\tau$ agree on $A_m$, which implies that they agree on all of $A$ by linearity.

Answer 3

1. The counit axiom implies that the coaction map $\rho: A \to A \otimes H$ is injective (it explicitly gives an inverse), hence the direct sum structure on $A \otimes H$ given by $H = \bigoplus_{ n \in \mathbb{Z}} k_{x^n} \otimes_k A$ pulls back via $\rho^{-1}$ to give a direct sum decomposition of $A$: $A = \bigoplus_{n \in \mathbb{Z}} \rho^{-1} (k_{x^n} \otimes A)$.

In fact, it is precisely in this step where you are using the coassociativity of the action. Direct sum decompositions do not generally pull back: For example, for any field $K$, the canonical direct sum decomposition $K^2 \cong K \oplus K$ does not pull back to a direct sum decomposition of the diagonal, since both intersections are empty! In order to show that in this case, the direct sum decomposition does pull back, you will need to use the coassociativity of the action.