# The Unapologetic Mathematician

## Coproduct Root Systems

We should also note that the category of root systems has binary (and thus finite) coproducts. They both start the same way: given root systems $\Phi$ and $\Phi'$ in inner-product spaces $V$ and $V'$, we take the direct sum $V\oplus V'$ of the vector spaces, which makes vectors from each vector space orthogonal to vectors from the other one.

The coproduct $\Phi\amalg\Phi'$ root system consists of the vectors of the form $(\alpha,0)$ for $\alpha\in\Phi$ and $(0,\alpha')$ for $\alpha'\in\Phi'$. Indeed, this collection is finite, spans $V\oplus V'$, and does not contain $(0,0)$. The only multiples of any given vector in $\Phi\amalg\Phi'$ are that vector and its negative. The reflection $\sigma_{(\alpha,0)}$ sends vectors coming from $\Phi$ to each other, and leaves vectors coming from $\Phi'$ fixed, and similarly for the reflection $\sigma_{(0,\alpha')}$. Finally,

\displaystyle\begin{aligned}(\beta,0)\rtimes(\alpha,0)=\beta\rtimes\alpha&\in\mathbb{Z}\\(0,\beta')\rtimes(0,\alpha')=\beta'\rtimes\alpha'&\in\mathbb{Z}\\(\beta,0)\rtimes(0,\alpha')=(0,\beta')\rtimes(\alpha,0)=0&\in\mathbb{Z}\end{aligned}

All this goes to show that $\Phi\amalg\Phi'$ actually is a root system. As a set, it’s the disjoint union of the two sets of roots.

As a coproduct, we do have the inclusion morphisms $\iota_1:\Phi\rightarrow\Phi\amalg\Phi'$ and $\iota_2:\Phi'\rightarrow\Phi\amalg\Phi'$, which are inherited from the direct sum of $V$ and $V'$. This satisfies the universal condition of a coproduct, since the direct sum does. Indeed, if $\Psi\subseteq U$ is another root system, and if $f:V\rightarrow U$ and $f':V'\rightarrow U$ are linear transformations sending $\Phi$ and $\Phi'$ into $\Psi$, respectively, then $(a,b)\mapsto f(a)+g(b)$ sends $\Phi\amalg\Phi'$ into $\Psi$, and is the unique such transformation compatible with the inclusions.

Interestingly, the Weyl group of the coproduct is the product $\mathcal{W}\times\mathcal{W}'$ of the Weyl groups. Indeed, for every generator $\sigma_\alpha$ of $\mathcal{W}$ and every generator $\sigma_{\alpha'}$ of $\mathcal{W}'$ we get a generator $\sigma_{(\alpha,0)}$. And the two families of generators commute with each other, because each one only acts on the one summand.

On the other hand, there are no product root systems in general! There is only one natural candidate for $\Phi\times\Phi'$ that would be compatible with the projections $\pi_1:V\oplus V'\rightarrow V$ and $\pi_2:V\oplus V'\rightarrow V'$. It’s made up of the points $(\alpha,\alpha')$ for $\alpha\in\Phi$ and $\alpha'\in\Phi'$. But now we must consider how the projections interact with reflections, and it isn’t very pretty.

The projections should act as intertwinors. Specifically, we should have

$\displaystyle\pi_1(\sigma_{(\alpha,\alpha')}(\beta,\beta'))=\sigma_{\pi_1(\alpha,\alpha')}(\pi_1(\beta,\beta'))=\sigma_\alpha(\beta)$

and similarly for the other projection. In other words

$\displaystyle\sigma_{(\alpha,\alpha')}(\beta,\beta')=(\sigma_\alpha(\beta),\sigma_{\alpha'}(\beta'))$

But this isn’t a reflection! Indeed, each reflection has determinant $-1$, and this is the composition of two reflections (one for each component) so it has determinant ${1}$. Thus it cannot be a reflection, and everything comes crashing down.

That all said, the Weyl group of the coproduct root system is the product of the two Weyl groups, and many people are mostly concerned with the Weyl group of symmetries anyway. And besides, the direct sum is just as much a product as it is a coproduct. And so people will often write $\Phi\times\Phi'$ even though it’s really not a product. I won’t write it like that here, but be warned that that notation is out there, lurking.

January 25, 2010 -

## 6 Comments »

1. Doesn’t the determinant argument leave open the possibility of (2n+1)-way products? A more general method might be to look at the dimensions of the various eigenspaces.

Comment by Chad | January 25, 2010 | Reply

2. Yes, Chad, the argument doesn’t cover odd-arity products, but a more technical argument like you suggest can cover those cases. Even without going down that road, though, it should be clear that we definitely don’t have all finite products, and so whatever remains (by your proposal: nothing) is far from the natural notion that coproducts are.

But it still doesn’t explain the ubiquitous use of $\times$ to denote coproduct root systems, even or odd. Indeed, it’s easy to find lists of two-dimensional root systems that include (if not start with) “$A_1\times A_1$“.

Comment by John Armstrong | January 25, 2010 | Reply

• Whoever said coproducts had to look like addition and products had to look like multiplication? In categories that behave more like $\text{Set}^{op}$ than $\text{Set}$, such as $\text{CRing}$, exactly the opposite is true.

Comment by Qiaochu Yuan | January 25, 2010 | Reply

3. Qiaochu, Of course they don’t have to look like that. But the use is so widespread that when one sees $\times$ one expects products.

More substantively, nobody really says what they mean by $\times$ in these contexts. It’s just taken as obvious. And they definitely don’t talk about coproducts (which they should!)

Comment by John Armstrong | January 25, 2010 | Reply

4. […] to each other, and together they span the whole of . Thus we can write , and thus is the coproduct […]

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5. […] we have no information about their relative lengths. This is to be expected, since when we form the coproduct of two root systems the roots from each side are orthogonal to each other, and we should have no […]

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