Grothendieck Rings and the Categorification of ℤ[M]

Let $M$$M$ be a monoid, view it as a discrete category $\mathcal{D}(M)$$\mathcal D(M)$. Let ${\mathsf{Vect}}_k^f$$\mathsf{Vect}_{k}^f$ be the category of finite dimensional vector space over $k$$k$.

Consider the following functor category $\mathcal{C}:={\mathsf{Func}}^f(\mathcal{D}(M),{\mathsf{Vect}}_k^f)$$\mathcal C:=\mathsf{Func}^f(\mathcal D(M),\mathsf{Vect}_{k}^f)$.

Here we only consider those functors with finite supports.

$$|\{m\in M:F(m)\ne 0\}|<\mathrm{\infty }$$

The functor category has a natural monoidal structure, which is given by

$$F\otimes G(m):=\underset{ab=m}{⨁}F(a){\otimes }_{k}G(b)$$

Easy to see that

$$F\cong G⟺\mathrm{\forall }m\in M,\dim F(m)=\dim G(m)$$

For a functor $F\in \mathcal{C}$$F\in\mathcal C$, we should define the dimension of $F$$F$ as:

$$\begin{array}{c}\dim F:=\sum _{m\in M}\dim F(m)\cdot m\in \mathbb{Z}[M]\end{array}$$

Then the $\dim$$\dim$ gives us an isomorphic between the Grothendieck Ring of $\mathcal{C}$$\mathcal C$ and $\mathbb{Z}[M]$$\mathbb Z[M]$.

$$\dim :K_0(\mathcal{C})\to \mathbb{Z}[M]$$

Hence $\mathcal{C}$$\mathcal C$ is a categorification of $\mathbb{Z}[M]$$\mathbb Z[M]$.

Now let us look at some combinatorics.

Let $M=(\mathbb{N},+)$$M=(\mathbb N,+)$. Then $\mathbb{Z}[M]\cong \mathbb{Z}[x]$$\mathbb Z[M]\cong \mathbb Z[x]$. We know that $\bigwedge (V)$$\bigwedge(V)$ is a $\mathbb{N}$$\mathbb N$ graded vector space, and $\bigwedge$$\bigwedge$ maps direct sums to tensor products.

We have

$$\bigwedge (k^n)\cong \underset{i=1}{\overset{n}{⨂}}\bigwedge (k)$$

Take $\dim$$\dim$ both sides we get

$$\dim \bigwedge (k^n)=\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})x^i=(1+x{)}^n=\dim \underset{i=1}{\overset{n}{⨂}}\bigwedge (k)$$

If we apply $\mathsf{Totaldim}:K_0(\mathcal{C})\to \mathbb{Z},[F]⟼\sum _{m\in M}\dim F(m)$$\mathsf{Totaldim}:K_0(\mathcal C)\to\mathbb Z,[F]\longmapsto\sum_{m\in M}\dim F(m)$, we get

$$\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})=2^n$$

Also, we have

$$\bigwedge (k^n\oplus k^m)\cong \bigwedge (k^n){\otimes }_k\bigwedge (k^m)$$

Hence

$$\stackrel{s}{\bigwedge }(k^n\oplus k^m)\cong \underset{i+j=s}{⨁}\stackrel{i}{\bigwedge }(k^n){\otimes }_k\stackrel{j}{\bigwedge }(k^m)$$

Take the ordinary dimension both sides we get

$$(\genfrac{}{}{0ex}{}{n+m}{s})=\sum _{i+j=s}(\genfrac{}{}{0ex}{}{n}{i})(\genfrac{}{}{0ex}{}{m}{j})$$

Equivalently we could consider

$${\dim }_m:K_0(\mathcal{C})\to \mathbb{Z},F⟼\dim F(m)$$