# The Unapologetic Mathematician

## The Zero Representation

Okay, this is going to sound pretty silly and trivial, but I’ve been grading today. There is one representation we always have for any group, called the zero representation $\mathbf{0}$.

Pretty obviously this is built on the unique zero-dimensional vector space $\mathbf{0}$. It shouldn’t be hard to convince yourself that $\mathrm{GL}(\mathbf{0})$ is the trivial group, and so any group $G$ has a unique homomorphism to this group. Thus there is a unique representation of $G$ on the vector space $\mathbf{0}$.

We should immediately ask: is this representation a zero object? Suppose we have a representation $\rho:G\rightarrow\mathrm{GL}(V)$. Then there is a unique arrow $0:V\rightarrow\mathbf{0}$ sending every vector $v\in V$ to $0\in\mathbf{0}$. Similarly, there is a unique arrow $0:\mathbf{0}\rightarrow V$ sending the only vector $0\in\mathbf{0}$ to the zero vector $0\in V$. It’s straightforward to show that these linear maps are intertwinors, and thus that the zero representation is indeed a zero object for the category of representations of $G$.

This is all well and good for groups, but what about representing an algebra $A$? This can only make sense if we allow rings without unit, which I only really mentioned back when I first defined a ring. This is because there’s only one endomorphism of the zero-dimensional vector space at all! The endomorphism algebra will consist of just the element $0:\mathbf{0}\rightarrow\mathbf{0}$, and a representation of $A$ has to be an algebra homomorphism to this non-unital algebra. Given this allowance, we do have the zero representation, and it’s a zero object just as for groups. It’s sort of convenient, so we’ll tacitly allow this one non-unital algebra to float around just so we can have our zero representation, even if we allow no other algebras without units.

December 8, 2008