Clarification about the basis of $gl(n, F)$ (Humphreys' book)

Question

From Humphreys - Introduction to Lie Algebras and Rep. Theory, page $2$:

"For reference, we write down the multiplication table for $gl(n, F)$ relative to the standard basis consisting of the matrices $e_{ij}$ (having $1$ in the $(i, j)$ position and $0$ elsewhere). Since $e_{ij}e_{kl} = \delta_{jk}e_{il}$ it follows that $$[e_{ij}, e_{kl}] = \delta_{jk}e_{il} - \delta_{li}e_{kj}$$

• it's not clear at all to me what does he mean with the standard basis consisting in the matrices $e_{ij}$ having $1$ at position $(i, j)$ and $0$ elsewhere.

For how it's written, it appreas to me that this basis is made by matrices with $1$ everywhere except when $i = j$. But this cannot be a basis.

Also, I know the Kronecker symbol $\delta$ but I don't get the meaning of the whole commutator operation.

Can someone please guide me with an example, like a $2\times 2$ matrix or a $3\times 3$ too?

NOTE: by $gl(n, F)$ it's meant the set of all $n\times n$ matrices over a field $F$, from the identification of $gl(V)$ with that set (for the reader who finds matrices easier to use).

Thank you!

Answer 1

In the $3 \times 3$ case, the element $e_{12}$ represents the $3 \times 3$ matrix with every entry equal zero but the entry $(1,2)$, which is equal to $1$:

$$ e_{12}=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$

Another example: $$ e_{21}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$

If you make the computation, you will see that $$ [e_{12},e_{21}]=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}+\begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}=e_{11}-e_{22}. $$

The formula $[e_{ij}, e_{kl}] = \delta_{jk}e_{il} - \delta_{li}e_{kj}$ just means that if $j=k$, then the term $e_{il}$ appears, if $j\neq k$ it doesn't. The same for $l$ and $i$. You can check that the example agrees with the formula.

Answer 2

I would prefer to use the notation $E_{ij}$ for the same object (your $e_{ij}$). We can think of this as an $n\times n$ matrix over $F$. Then, it has $n^2$ entries, organised into rows and columns. Specifying a row and a column gives you a unique entry in the matrix. Thus, it makes sense to say that $E_{ij}$ is the matrix with a $1$ for the $i,j$'th entry, and $0$ everywhere else. In $3$ dimensions for example, $$ E_{1,2}=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 &0 & 0 \end{pmatrix}. $$ Considering $gl(n,F)$ as a $F$-vector space of all $n\times n$ matrices over $F$, it is clear that the set $\lbrace E_{ij}\rbrace_{1\leq i,j\leq n}$ is a basis. Now by matrix multiplication, it is easy to verify Humphreys' calculations.

Answer 3

It looks like your describing the standard $F$-basis which the in case of $2\times 2$ is given by $$ e_{1,1}= \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right), \, e_{1,2}= \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right), \, e_{2,1}= \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right), \, \text{ and } e_{2,2}= \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right).$$

Then you have the equality $e_{i,j} \, e_{k,l} = \begin{cases} e_{i,l} & : j=k \\ 0 &: \text{else}\end{cases}$.

As in our case the Lie-Bracket is given by the commutator $[a,b] = ab - ba$,
the above equation is exactly what you get for the $e_{i,j}$.