How do I prove a set is a root system?

Question

I will start by setting up the problem then I will ask my question regarding it.

Problem:

Let $E=\{x=\sum_{i=1}^{n+1} x_{i}\epsilon_{i} \in \mathbb{R}^{n+1}|\sum_{i=1}^{n+1}x_{i}=0 \}$, where $\{\epsilon_{1},\dots,\epsilon_{n+1}\}$, is the standard basis of $\mathbb{R}^{n+1}$ with $(\epsilon_{i},\epsilon_{j})= \delta_{i,j}$.

Let $R=\{\epsilon_{i}-\epsilon_{j}| 1\le i \neq j \ge n+1\}$.

Show that $R$ is a root system.

Question:

I understand how a root system works and the concepts surrounding it however I am unsure how to deal with it in this formulation. I am given solutions to it however they simply state that it is obvious $\{\epsilon_{1}-\epsilon_{2}, \dots , \epsilon_{n}-\epsilon_{n+1}\}$ is a basis, $0$ is not contained in $R$ and the only multiples of $R$ are its negatives. It then goes on to check the third axiom as follows: Axiom 3

(I couldn't for the life of me format the pictured information in Latex - sorry!).

Then something similar is done for axiom 4.

None of this is obvious to me even if the solution says it is. How do these values fulfil the axioms? What is being done in axiom 3 to discern the values? I know why they fulfil the requirement but not how they got there.

Thanks for your time!

Answer 1

Technically, a set alone cannot be a root system; all standard definitions of a root system contain some extra structure, like an ambient vector space, certain reflections or dual roots or a scalar product etc.

In this case, I assume your source takes for granted the standard scalar product on the vector space $\mathbb R^{n+1}$ or its restriction to $E$ as ambient vector space of $R$ though.

But then, maybe it's just a notational issue. Are you clear that $\epsilon_1-\epsilon_2$ is the vector $\pmatrix{1\\-1\\0\\0\\\vdots \\0}$, $\epsilon_2-\epsilon_3 = \pmatrix{0\\1\\-1\\0\\\vdots \\0}$ etc, and more generally $\epsilon_i - \epsilon_j$ is the one with a $1$ at position $i$ and a $-1$ at position $j$?

Then I'd think it is clear that the zero vector $0$ is not any one of these elements in $R$ which, as just said, have a $1$ somewhere and a $-1$ somewhere else. It is also clear that the only multiples of $\epsilon_i - \epsilon_j$ which are again of the form $\epsilon_k - \epsilon_l$ are the ones with $\{k,l\} = \{i,j\}$ i.e. $\pm (\epsilon_i-\epsilon_j)$.

It should be, if not yet obvious, then easy to show (in the following order) with standard linear algebra, that

1. the vector space $E$, which contains $R$, has dimension $n$ (e.g. it is the kernel of the non-zero linear form $l : \mathbb R^{n+1} \rightarrow \mathbb R, \pmatrix{x_1\\x_2\\\vdots \\x_{n+1}} \mapsto \sum x_i$)

2. the set $R$ generates $E$ as vector space

3. the elements $\epsilon_1 - \epsilon_2, \epsilon_2-\epsilon_3, ..., \epsilon_{n}-\epsilon_{n+1}$ are linearly independent and hence a vector space basis of $E$

4. actually, for a given $\epsilon_i -\epsilon_j$ with $j \ge i+1$, it should be straightforward to write $\epsilon_i - \epsilon_j$ as a linear combination of $\epsilon_i-\epsilon_{i+1}, \epsilon_{i+1}-\epsilon_{i+2}, ..., \epsilon_{j-1}-\epsilon_j$ (hint: add them); and for the case $i \ge j+1$, the negative sign was invented.

Finally, to get the values in the image you linked, well the standard scalar product is bilinear, i.e.

$$\langle \epsilon_i -\epsilon_j, \epsilon_k-\epsilon_l \rangle = \langle \epsilon_i, \epsilon_k \rangle - \langle \epsilon_i, \epsilon_l\rangle - \langle \epsilon_j, \epsilon_k \rangle + \langle \epsilon_j, \epsilon_l \rangle = \delta_{i,k}-\delta_{i,l}-\delta_{j,k}+\delta_{j,l}$$

and if you know the definition of the Kronecker delta you should see why that gives the values in all cases of identities among the $i,j,k,l$ as claimed. Of course you can double check scalar product examples like $\langle \pmatrix{1\\-1\\0\\0\\\vdots \\0}, \pmatrix{0\\1\\-1\\0\\\vdots \\0} \rangle = -1$ (here $j=k=2, i=1 \neq l=3$) by hand if that is psychologically easier at first, but I hope you quickly get to see the shorter notation is more practical.