The Stacks project

12.16 Graded objects

We make the following definition.

Definition 12.16.1. Let $\mathcal{A}$ be an additive category. The category of graded objects of $\mathcal{A}$, denoted $\text{Gr}(\mathcal{A})$, is the category with

  1. objects $A = (A^ i)$ are families of objects $A^ i$, $i \in \mathbf{Z}$ of objects of $\mathcal{A}$, and

  2. morphisms $f : A = (A^ i) \to B = (B^ i)$ are families of morphisms $f^ i : A^ i \to B^ i$ of $\mathcal{A}$.

If $\mathcal{A}$ has countable direct sums, then we can associate to an object $A = (A^ i)$ of $\text{Gr}(\mathcal{A})$ the object

\[ A = \bigoplus \nolimits _{i \in \mathbf{Z}} A^ i \]

and set $k^ iA = A^ i$. In this case $\text{Gr}(\mathcal{A})$ is equivalent to the category of pairs $(A, k)$ consisting of an object $A$ of $\mathcal{A}$ and a direct sum decomposition

\[ A = \bigoplus \nolimits _{i \in \mathbf{Z}} k^ iA \]

by direct summands indexed by $\mathbf{Z}$ and a morphism $(A, k) \to (B, k)$ of such objects is given by a morphism $\varphi : A \to B$ of $\mathcal{A}$ such that $\varphi (k^ iA) \subset k^ iB$ for all $i \in \mathbf{Z}$. Whenever our additive category $\mathcal{A}$ has countable direct sums we will use this equivalence without further mention.

However, with our definitions an additive or abelian category does not necessarily have all (countable) direct sums. In this case our definition still makes sense. For example, if $\mathcal{A} = \text{Vect}_ k$ is the category of finite dimensional vector spaces over a field $k$, then $\text{Gr}(\text{Vect}_ k)$ is the category of vector spaces with a given gradation all of whose graded pieces are finite dimensional, and not the category of finite dimensional vector spaces with a given graduation.

Lemma 12.16.2. Let $\mathcal{A}$ be an abelian category. The category of graded objects $\text{Gr}(\mathcal{A})$ is abelian.

Proof. Let $f : A = (A^ i) \to B = (B^ i)$ be a morphism of graded objects of $\mathcal{A}$ given by morphisms $f^ i : A^ i \to B^ i$ of $\mathcal{A}$. Then we have $\mathop{\mathrm{Ker}}(f) = (\mathop{\mathrm{Ker}}(f^ i))$ and $\mathop{\mathrm{Coker}}(f) = (\mathop{\mathrm{Coker}}(f^ i))$ in the category $\text{Gr}(\mathcal{A})$. Since we have $\mathop{\mathrm{Im}}= \mathop{\mathrm{Coim}}$ in $\mathcal{A}$ we see the same thing holds in $\text{Gr}(\mathcal{A})$. $\square$

Remark 12.16.3 (Warning). There are abelian categories $\mathcal{A}$ having countable direct sums but where countable direct sums are not exact. An example is the opposite of the category of abelian sheaves on $\mathbf{R}$. Namely, the category of abelian sheaves on $\mathbf{R}$ has countable products, but countable products are not exact. For such a category the functor $\text{Gr}(\mathcal{A}) \to \mathcal{A}$, $(A^ i) \mapsto \bigoplus A^ i$ described above is not exact. It is still true that $\text{Gr}(\mathcal{A})$ is equivalent to the category of graded objects $(A, k)$ of $\mathcal{A}$, but the kernel in the category of graded objects of a map $\varphi : (A, k) \to (B, k)$ is not equal to $\mathop{\mathrm{Ker}}(\varphi )$ endowed with a direct sum decomposition, but rather it is the direct sum of the kernels of the maps $k^ iA \to k^ iB$.

Definition 12.16.4. Let $\mathcal{A}$ be an additive category. If $A = (A^ i)$ is a graded object, then the $k$th shift $A[k]$ is the graded object with $A[k]^ i = A^{k + i}$.

If $A$ and $B$ are graded objects of $\mathcal{A}$, then we have

12.16.4.1
\begin{equation} \label{homology-equation-hom-into-shift} \mathop{\mathrm{Hom}}\nolimits _{\text{Gr}(\mathcal{A})}(A, B[k]) = \mathop{\mathrm{Hom}}\nolimits _{\text{Gr}(\mathcal{A})}(A[-k], B) \end{equation}

and an element of this group is sometimes called a map of graded objects homogeneous of degree $k$.

Given any set $G$ we can define $G$-graded objects of $\mathcal{A}$ as the category whose objects are $A = (A^ g)_{g \in G}$ families of objects parametrized by elements of $G$. Morphisms $f : A \to B$ are defined as families of maps $f^ g : A^ g \to B^ g$ where $g$ runs over the elements of $G$. If $G$ is an abelian group, then we can (unambiguously) define shift functors $[g]$ on the category of $G$-graded objects by the rule $(A[g])^{g_0} = A^{g + g_0}$. A particular case of this type of construction is when $G = \mathbf{Z} \times \mathbf{Z}$. In this case the objects of the category are called bigraded objects of $\mathcal{A}$. The $(p, q)$ component of a bigraded object $A$ is usually denoted $A^{p, q}$. For $(a, b) \in \mathbf{Z} \times \mathbf{Z}$ we write $A[a, b]$ in stead of $A[(a, b)]$. A morphism $A \to A[a, b]$ is sometimes called a map of bidegree $(a, b)$.


Comments (3)

Comment #1291 by Tim Porter on

Line 7: typo: associate (i.e. delete the d)

I think your entry is wrong. cannot be equivalent to the category of objects in with countable direct sum decompositions in the case that countable direct products are not exact. If it were then kernels would correspond. The point surely is that you have not made precise what this second category actually is. The notation is not explained in any detail here but if the are the projections you also need the 'injections'. With those, I am not sure that the point you are making still is valid as countable products in need not be the same as countable products in the category of 'objects with direct sum decomposition'.

As your page is exposition of an elementary idea, this level of detailed comment may not be needed, but at present the point is blurred because you seem to say that there are equivalent categories in which kernels do not agree, which is clearly not what is intended.

Generally, thanks for the project. It is very useful.

Comment #1303 by on

Hi Tim! Thanks for the typo which I fixed here.

Your complaint that we do not have a clear definition of what is the category of pairs is certainly valid.

Assume that has all countable direct sums. Say for a family of objects . Then there are canonical maps whose composition is the identity. Namely, and we can use Lemma 12.3.4.

So, we can think of the category of pairs as the category of objects which are endowed with a family of subobjects for such that the induced morphism is an isomorphism. Morphisms are morphisms such that .

With this definition I think what is said is correct. OK?

Comment #9741 by luciano on

I think "graduation" should be "gradation". I'm not sure that "graduation" is wrong, many people use it too, but since the page also uses "gradation" I think it's better to stick with that one


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