A What is the definition of quotient Lie algebra?

[HDB1]

TL;DR Summary

quotient Lie algebra

Please, in the definition of quotient Lie algebra

If $I$ is an ideal of $\mathfrak{g}$, then the vector space $\mathfrak{g} / I$ with the bracket defined by:
$$[x+I, y+I]=[x, y]+I, for all x, y \in \mathfrak{g}$$,
is a Lie algebra called the quotient Lie algebra of $\mathfrak{g}$ by $I$.

Since, $\mathfrak{s l}_2(\mathbb{K}) \text { is an ideal of the lie algebra } \mathfrak{g l}_2(\mathbb{K}) .$

could we define quotient Lie algebra here, is it the function from $\mathfrak{g l}_2$ to $\mathfrak{g l}_2 / \mathfrak{s l}_2$,?
if yes, how we get this element, $x+I =x + \mathfrak{s l}_2, x \in \mathfrak{g l}_2$

Thank you so much,

 

[HDB1]

Dear @fresh_42 , if you could help, I would appreciate it,

 

[fresh_42]

TL;DR Summary: quotient Lie algebra

Please, in the definition of quotient Lie algebra

If $I$ is an ideal of $\mathfrak{g}$, then the vector space $\mathfrak{g} / I$ with the bracket defined by:
$$[x+I, y+I]=[x, y]+I, \forall x, y \in \mathfrak{g}$$,
is a Lie algebra called the quotient Lie algebra of $\mathfrak{g}$ by $I$.

Since, $\mathfrak{s l}_2(\mathbb{K}) \text { is an ideal of the lie algebra } \mathfrak{g l}_2(\mathbb{K}) .$

could we define quotient Lie algebra here, is it the function from $\mathfrak{g l}_2$ to $\mathfrak{g l}_2 / \mathfrak{s l}_2$,?
if yes, how we get this element, $x+I =x + \mathfrak{s l}_2, x \in \mathfrak{g l}_2$

Thank you so much,

Yes, this is true.

We can split $\mathfrak{gl}(2)=\mathfrak{sl}(2)\oplus \mathfrak{Z}(\mathfrak{sl})(2)=\mathfrak{sl}(2)\oplus \mathbb{K}\cdot \begin{pmatrix}1&0\\0&1\end{pmatrix}.$ It is a direct sum, i.e. both are ideals in $\mathfrak{gl}(2).$ Hence we can build both quotients ($\mathbb{1}$ is the identity matrix)
\begin{align*}
\mathfrak{gl}(2) &\twoheadrightarrow \mathfrak{gl}(2)/\mathfrak{sl}(2)\cong \mathbb{K}\\[16pt]
\mathfrak{gl}(2) &\twoheadrightarrow \mathfrak{gl}(2)/ \mathbb{K}\cdot \mathbb{1}\cong \mathfrak{sl}(2)
\end{align*}
The two projection mappings are given by the following decomposition
\begin{align*}
\begin{pmatrix}a&b\\c&d\end{pmatrix}\longmapsto \begin{pmatrix}a&b\\c&d\end{pmatrix}+\mathfrak{sl}(2)=&\begin{pmatrix}\dfrac{d+a}{2}&0\\0&\dfrac{d+a}{2}\end{pmatrix}+\underbrace{\begin{pmatrix}\dfrac{a-d}{2}&b\\c&\dfrac{d-a}{2}\end{pmatrix}}_{\in \mathfrak{sl}(2)} +\mathfrak{sl}(2) \\
&=\underbrace{\begin{pmatrix}\dfrac{d+a}{2}&0\\0&\dfrac{d+a}{2}\end{pmatrix}}_{= \frac{a+d}{2}\cdot \mathbb{1}}+\mathfrak{sl}(2)\in \mathbb{K}\cdot \mathbb{1}+\mathfrak{sl}(2)
\end{align*}
The general linear algebras $\mathfrak{gl}(V)$ are the direct product of its ideals $[\mathfrak{gl}(V),\mathfrak{gl}(V)]=\mathfrak{sl}(V)$ and its one-dimensional center $\mathfrak{Z}(\mathfrak{gl}(V))\cong \mathbb{K}$ of multiples of the identity matrix:
$$
\mathfrak{gl}(V) \cong \mathfrak{sl}(V) \times \mathbb{K}
$$

Note: $\times$ or $\oplus$ are the same here. I prefer $\times$ since it signals that both factors are ideals, and $\oplus$ is primarily a direct sum of vector spaces.

 

[HDB1]

\begin{align*}
\mathfrak{gl}(2) &\twoheadrightarrow \mathfrak{gl}(2)/\mathfrak{sl}(2)\cong \mathbb{K}\\[16pt]
\mathfrak{gl}(2) &\twoheadrightarrow \mathfrak{gl}(2)/ \mathbb{K}\cdot \mathbb{1}\cong \mathfrak{sl}(2)
\end{align*}

Thank you so much, I really appreciate your time and help,

##\mathfrak{gl}(2)=\mathfrak{sl}(2)\oplus \mathfrak{Z}(\mathfrak{sl})(2)

here please, you the center of $\mathfrak{g l}(2)$ or $\mathfrak{s l}(2)$please, I have a question, here, how you get the isomorophism here:

could you please, give me example how we compute: $\mathfrak{g l}(2) / \mathfrak{s l}(2)$

is the outcome will be in $\mathfrak{g l}(2)$? please.
I mean in general , is the factor algebra of $\mathfrak{g l}(2) / \mathfrak{s l}(2)$ will give us an matrix in $\mathfrak{g l}(2)$please, here: how you get it:

$[\mathfrak{gl}(V),\mathfrak{gl}(V)]=\mathfrak{sl}(V)$

I just use tow matrices form $\mathfrak{g l}(2)$ and compute their commutator and get a matrix in ## / \mathfrak{s l}(2)
is that enough? please,

$\mathfrak{Z}(\mathfrak{gl}(V))\cong \mathbb{K}$

I am so sorry, but please, could you tell me how you get the center?

 

[fresh_42]

Thank you so much, I really appreciate your time and help,

please, I have a question, here, how you get the isomorophism here:

\begin{align*}
\mathfrak{gl}(2) &\twoheadrightarrow \mathfrak{gl}(2)/\mathfrak{sl}(2)\cong \mathbb{K}\\[16pt]
\mathfrak{gl}(2) &\twoheadrightarrow \mathfrak{gl}(2)/ \mathbb{K}\cdot \mathbb{1}\cong \mathfrak{sl}(2)
\end{align*}

could you please, give me example how we compute: $\mathfrak{g l}(2) / \mathfrak{s l}(2)$

Quotient spaces are sets of equivalence classes. Given $A/B$ we consider its elements as $a+B.$ This means that $a+B =a'+B \Longleftrightarrow a-a' \in B.$

The easiest way to get used to it is to consider, say $\mathbb{Z}/3\cdot \mathbb{Z}.$ It consists of elements $z_0+3\cdot \mathbb{Z}.$ So the elements themselves are sets, represented by $z_0.$ $3\mathbb{Z}=\{\ldots,-6,-3,0,3,6,9,12,\ldots\}.$ In $\mathbb{Z}/3\cdot \mathbb{Z}$ we identify
$$
a+3\mathbb{Z}=a'+3\mathbb{Z} \Longleftrightarrow a-a' \in 3\mathbb{Z}
$$
This results in three elements:
\begin{align*}
\bar 0=0+3\mathbb{Z}&=\{\ldots,-6,-3,0,3,6,9,12,\ldots\}\\
\bar 1=1+3\mathbb{Z}&=\{\ldots,-5,-2,1,4,7,10,13,\ldots\}\\
\bar 2=2+3\mathbb{Z}&=\{\ldots,-4,-1,2,5,8,11,14,\ldots\}
\end{align*}
Every integer is in one of these sets and represented by one of the elements $\bar 0\, , \,\bar 1,\bar 2.$ So $$
\mathbb{Z}/3\mathbb{Z}=\mathbb{Z}_3=\mathbb{F}_3=\{\bar 0\, , \,\bar 1,\bar 2\}=\{[0],[1],[2]\}=\{0,1,2\}
$$
The letter $\mathbb{F}$ signals that $\mathbb{Z}_3$ is a field (of characteristic $3$) so it is sometimes written $\mathbb{F}_3$ and the set $\{0,1,2\}$ comes from lazy people who do not want to carry the bar, or sometimes brackets. The brackets here have nothing to do with Lie algebras. They are only a possible notation for equivalence classes. But as soon as it is clear that we speak about a quotient, i.e. equivalence classes as elements, as soon people simplify notation by just writing, here $0,1,2$ although they mean $0+3\mathbb{Z}\, , \,1+3\mathbb{Z}\, , \,2+3\mathbb{Z}.$

This is the construction plan behind quotients. In this example, we have in $0,1,2$ the possible remainders by an integer division by $3$. And these three numbers (with $2+1=0\, , \,2\cdot\otimes =1$) form a field. The set under the quotient is the new zero. Here all numbers divisible by $3$ become zero.

The quotients for vector spaces or rings or algebras or Lie algebras work accordingly.

Consider an arbitrary matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in \mathfrak{gl}(2).$ Then we write it as
$$
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\underbrace{\begin{pmatrix}\dfrac{a+d}{2}&0\\0&\dfrac{a+d}{2}\end{pmatrix}}_{=\lambda \mathbb{1}}
+ \underbrace{\begin{pmatrix}\dfrac{a-d}{2}&b\\c&\dfrac{d-a}{2}\end{pmatrix}}_{=\;X\;\in \mathfrak{sl}(2)}
$$
The first part is a multiple of the identity matrix $\mathbb{1}$, and the second part is in $\mathfrak{sl}(2).$

A) The multiples of the identity matrix, $\mathbb{K}\cdot \mathbb{1}$ commute with all other matrices, so they are in the center or $\mathfrak{gl}(2).$ They are in fact the entire center (Schur's lemma, I think). The center of a Lie algebra is an ideal. So we can not only build $\mathfrak{gl}(2)/\mathbb{K}\cdot \mathbb{1} =\mathfrak{gl}(2)/\mathfrak{Z}(\mathfrak{gl}(2)),$ the ideal property also guarantees that this quotient is again a Lie algebra. (Try to prove: If $I\triangleleft L$ is an ideal, then $L/I=\{a+I\,|\,a\in L\}$ is again a Lie algebra.) Here we have
$$
\bar A = A+\{\lambda \cdot \mathbb{1}\,|\,\lambda \in \mathbb{K}\}= X
$$
The first part of the decomposition of $A=\lambda\cdot \mathbb{1}+X$ is swallowed in $\{\lambda \cdot \mathbb{1}\,|\,\lambda \in \mathbb{K}\}$ which is nothing else as the center of $\mathfrak{gl}(2),$ or a copy of the scalar field $\mathbb{K}$ because we have only one scalar as a parameter, it is a one-dimensional vector space:
$$
\mathfrak{gl}(2)/\mathfrak{Z}(\mathfrak{gl}(2))=\mathfrak{gl}(2)/\mathbb{K}\cdot \mathbb{1}\cong \{X\} =\mathfrak{sl}(2)
$$
The isomorphism is $\varphi (A)=\bar A=X+\mathfrak{Z}(\mathfrak{gl}(2))$ and since nobody wants to write this, we simply say $\varphi (A)=X$ where $X$ is the part of $A$ that is in $\mathfrak{sl}(2).$

B) The second possibility is accordingly. We have
$$
[\mathfrak{gl}(2),\mathfrak{gl}(2)]=\mathfrak{sl}(2)
$$
so $\mathfrak{sl}(2)\triangleleft \mathfrak{gl}(2)$ is also an ideal in $\mathfrak{gl}(2).$ Therefore
$$
\mathfrak{gl}(2)/\mathfrak{sl}(2) \cong \mathbb{K}
$$
is again a Lie algebra. This time, we make $\mathfrak{sl}(2)$ our new zero in the quotient space. It swallows the parts of $A$ that is in $\mathfrak{sl}(2),$ i.e. it swallows $X$. The isomorphism is thus $\varphi (A)=\bar A=\lambda \cdot\mathbb{1}+\mathfrak{sl}(2),$ and again if we strip all redundant notation, we get $\varphi (A)=\lambda \in \mathbb{K}.$ Strictly speaking, it would have been $\mathbb{K}\cdot \mathbb{1}$ but this is the same as just writing $\mathbb{K}.$

Note: The decomposition of $A$ also shows, that $\mathfrak{gl}(2)=\mathbb{K}\cdot \mathbb{1}\oplus \mathfrak{sl}(2)$ as mere vector spaces, just by the addition of $A=\lambda \cdot \mathbb{1}+X.$ This allows us to calculate
\begin{align*}
[\mathfrak{gl}(2),\mathfrak{gl}(2)]&=[\mathbb{K}\cdot \mathbb{1}\oplus \mathfrak{sl}(2),\mathbb{K}\cdot \mathbb{1}\oplus \mathfrak{sl}(2)]\\
&=\underbrace{[\mathbb{K}\cdot \mathbb{1},\mathbb{K}\cdot \mathbb{1}]}_{=\{0\}}+
\underbrace{[\mathbb{K}\cdot \mathbb{1},\mathfrak{sl}(2)]}_{=\{0\}}+\underbrace{[\mathfrak{sl}(2),\mathbb{K}\cdot \mathbb{1}]}_{=\{0\}}+[\mathfrak{sl}(2),\mathfrak{sl}(2)]\\
&=[\mathfrak{sl}(2),\mathfrak{sl}(2)]=\mathfrak{sl}(2)
\end{align*}

 

[fresh_42]

I just use tow matrices form gl(2) and compute their commutator and get a matrix in ## / \mathfrak{s l}(2)
is that enough? please,

Yes. You only have to show that $\operatorname{trace}[A,B]=\operatorname{trace}(A\cdot B-B\cdot A)=\operatorname{trace}(A\cdot B)-\operatorname{trace}(B\cdot A)=0.$

 

[HDB1]

$$
\bar A = A+\{\lambda \cdot \mathbb{1}\,|\,\lambda \in \mathbb{K}\}= X
$$
The first part of the decomposition of $A=\lambda\cdot \mathbb{1}+X$ is swallowed in $\{\lambda \cdot \mathbb{1}\,|\,\lambda \in \mathbb{K}\}$ which is nothing else as the center of $\mathfrak{gl}(2),$ or a copy of the scalar field $\mathbb{K}$ because we have only one scalar as a parameter, it is a one-dimensional vector space:
$$
\mathfrak{gl}(2)/\mathfrak{Z}(\mathfrak{gl}(2))=\mathfrak{gl}(2)/\mathbb{K}\cdot \mathbb{1}\cong \{X\} =\mathfrak{sl}(2)
$$
The isomorphism is $\varphi (A)=\bar A=X+\mathfrak{Z}(\mathfrak{gl}(2))$ and since nobody wants to write this, we simply say $\varphi (A)=X$ where $X$ is the part of $A$ that is in $\mathfrak{sl}(2).$

B) The second possibility is accordingly. We have
$$
[\mathfrak{gl}(2),\mathfrak{gl}(2)]=\mathfrak{sl}(2)
$$
so $\mathfrak{sl}(2)\triangleleft \mathfrak{gl}(2)$ is also an ideal in $\mathfrak{gl}(2).$ Therefore
$$
\mathfrak{gl}(2)/\mathfrak{sl}(2) \cong \mathbb{K}
$$
is again a Lie algebra. This time, we make $\mathfrak{sl}(2)$ our new zero in the quotient space. It swallows the parts of $A$ that is in $\mathfrak{sl}(2),$ i.e. it swallows $X$. The isomorphism is thus $\varphi (A)=\bar A=\lambda \cdot\mathbb{1}+\mathfrak{sl}(2),$ and again if we strip all redundant notation, we get $\varphi (A)=\lambda \in \mathbb{K}.$ Strictly speaking, it would have been $\mathbb{K}\cdot \mathbb{1}$ but this is the same as just writing $\mathbb{K}.$

Dear @fresh_42, thank you so much for your help and time,
please, if you do not mind, could you please, give me an example with number about above, I did not get the idea well,

 

[fresh_42]

Dear @fresh_42, thank you so much for your help and time,
please, if you do not mind, could you please, give me an example with number about above, I did not get the idea well,

You should actually start with integers and the example $\mathbb{Z}/3\mathbb{Z}=\mathbb{Z}_3=\{0,1,2\}$ I gave you. Try to prove that this set ...
\begin{align*}
&\left\{ 0+\{\ldots,-6,-3,0,3,6,\ldots\} \right\} \; , \; \left\{ 1+\{ \ldots,-6,-3,0,3,6,\ldots\} \right\} \; , \; \left\{ 2+\{ \ldots,-6,-3,0,3,6,\ldots \} \right\} \\
&=\left\{ \underbrace{\{ \ldots,-6,-3,0,3,6,\ldots \}}_{=0}\; , \;\underbrace{\{ \ldots,-5,-4,1,4,5,\ldots \}}_{=1} \; , \; \underbrace{\{ \ldots,-4,-1,2,5,7,\ldots \}}_{=2} \right\}
\end{align*}
... of three sets, each one an element of $\mathbb{Z}/3\mathbb{Z}$ form a field.

A.) $\mathfrak{gl}(2) / \mathbb{K} \cong \mathfrak{sl}(2)$

The elements of $\mathfrak{gl}(2) / \mathbb{K}$ are of the form
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix} + \mathbb{K}\cdot \mathbb{1}=\begin{pmatrix}a&b\\c&d\end{pmatrix}+\left\{\begin{pmatrix}\lambda &0\\0&\lambda \end{pmatrix}\, : \,\lambda \in \mathbb{K}\right\}
$$
If we have $\begin{pmatrix}17&-3\\11&-1\end{pmatrix}\in \mathfrak{gl}(2)$ then we write it as a sum as follows:
\begin{align*}
\underbrace{\begin{pmatrix}17&-3\\11&-1\end{pmatrix}}_{=X}&=\underbrace{\begin{pmatrix}\dfrac{17+1}{2}&-3\\11&\dfrac{-1-17}{2}\end{pmatrix}}_{=\bar X}+\underbrace{\begin{pmatrix}\dfrac{17-1}{2}&0\\0&\dfrac{17-1}{2}\end{pmatrix}}_{=8\cdot \mathbb{1}}\\
&=\bar X+8\cdot\mathbb{1}\in \mathfrak{sl}(2) + \mathbb{K}\cdot \mathbb{1}
\end{align*}
If we pass from $\mathfrak{gl}(2)$ to $\mathfrak{gl}(2) / \mathbb{K}$ then we consider all multiples of the identity matrix as (new) zero. So $X=\begin{pmatrix}17&-3\\11&-1\end{pmatrix}$ becomes (= is represented by) $\bar X.$ The part $\lambda \cdot \mathbb{1}$ is within $\mathbb{K}\cdot \mathbb{1}\cong \mathbb{K}$ and considered zero. The isomorphism maps
$$
\mathfrak{gl}(2)/\mathbb{K}\ni X+\mathbb{K} = \begin{pmatrix}17&-3\\11&-1\end{pmatrix}+\mathbb{K}\cdot \mathbb{1}\longmapsto
\begin{pmatrix}9&-3\\11&-9\end{pmatrix}=\bar X \in \mathfrak{sl}(2)
$$
We extracted $\begin{pmatrix}8&0\\0&8\end{pmatrix}$ from $X$ and pushed it into our new zero $\begin{pmatrix}8&0\\0&8\end{pmatrix}\in \bar 0=0+\mathbb{K}\cdot \mathbb{1}$ so only $\bar X \in \mathfrak{sl}(2)$ was left. In short:

A matrix from $\mathfrak{gl}(V)=\mathfrak{gl}(n)$ modulo its center $Z(\mathfrak{gl}(n))=\mathbb{K}\cdot \mathbb{1}\cong \mathbb{K}$ is in $\mathfrak{sl}(V)=\mathfrak{sl}(n).$B.) $\mathfrak{gl}(2) / \mathfrak{sl}(2) \cong \mathbb{K}$

This is the same as before with the roles of $\mathbb{K}\cdot \mathbb{1}$ and $\mathfrak{sl}(2)$ exchanged. This time we keep the diagonal matrix $\begin{pmatrix}8&0\\0&8\end{pmatrix}$ and push the remaining part $\begin{pmatrix}9&-3\\11&-9\end{pmatrix}$ into the new zero here, which is $\mathfrak{sl}(2)$. It is not the center of $\mathfrak{gl}(2)$ but the derived algebra $[\mathfrak{gl}(2)\, , \,\mathfrak{gl}(2)]$ which is also an ideal. Hence this isomorphism maps
$$
\mathfrak{gl}(2)/\mathfrak{sl}(2) \ni X+\mathfrak{sl}(2) = \begin{pmatrix}17&-3\\11&-1\end{pmatrix}+\mathfrak{sl}(2)\longmapsto
\begin{pmatrix}8&0\\0&8\end{pmatrix}=\lambda \cdot\mathbb{1} \in \mathbb{K}\cdot \mathbb{1}\cong \mathbb{K}
$$
We extracted $\begin{pmatrix}9&-3\\11&-9\end{pmatrix}$ from $X$ and #pushed it into our new zero $\begin{pmatrix}9&-3\\11&-9\end{pmatrix}\in \bar 0=0+\mathfrak{sl}(2)$ so only $\lambda \cdot \mathbb{1} \in \mathbb{K}\cdot \mathbb{1}\cong \mathbb{K}$ was left. In short:

A matrix from $\mathfrak{gl}(V)=\mathfrak{gl}(n)$ modulo its derived algebra $[\mathfrak{gl}(n)),\mathfrak{gl}(n))]=\mathfrak{sl}(V)=\mathfrak{sl}(n))$ is in $\mathbb{K}\cdot \mathbb{1}\cong \mathbb{K}.$

Note: This is an important concept, and it can easily be practiced with integers. Try to figure out why $\mathbb{Z}/5\mathbb{Z}=\mathbb{Z}_5$ is a field, whereas $\mathbb{Z}/4\mathbb{Z}=\mathbb{Z}_4$ is not. With $\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}_n$ we mean the set of remainders by division by $n$. If we have an arbitrary number $z\in \mathbb{Z}$ then we can write it as $z=q\cdot n+r$ with a remainder $r\in \{0,1,2,\ldots,n-1\}.$ Those remainders are the new elements in $\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}_n$. The elements are only unique up to are multiple of $n.$ Say $n=5$ and $z=7.$ Then $7=1\cdot 5 +2$ and $7$ is represented by $2$. But nobody prevents us from writing $7=11^2\cdot 5 -598$ and $7$ is now represented by $-598.$ However, $2-(-598)$ is a multiple of $5$ and thus regarded as the same number in the quotient ring $\mathbb{Z}/5\mathbb{Z}=\mathbb{Z}_5.$ The elements $0,1,2,3,4$ represent the sets $0+5\mathbb{Z},1+5\mathbb{Z},2+5\mathbb{Z},3+5\mathbb{Z},4+5\mathbb{Z}$ and $2\in 2+5\mathbb{Z}$ as is $-598 \in 2+5\mathbb{Z}.$

If you get used to this modular arithmetic then the concept of quotients will be much easier to understand.

 

[HDB1]

Thank you so much,@fresh_42 ,
I am so sorry for bothering you,

please, I have a question, in $\mathfrak{s l}(V)=\mathfrak{s l}(n))$, especially: when $n=2$ how we get the first direction, I just focus in $\mathfrak{s l}(2))$ but I did not use any example in $\mathfrak{s l}(V)$, I know it a map from $V$ to $V$, but how it equals $\mathfrak{s l}(n))$,
we have to compute the trace of this map, but please, if yes, do you have an example of it?

thanks in advance,

 

[fresh_42]

Thank you so much,@fresh_42 ,
I am so sorry for bothering you,

please, I have a question, in $\mathfrak{s l}(V)=\mathfrak{s l}(n))$, especially: when $n=2$ how we get the first direction, I just focus in $\mathfrak{s l}(2))$ but I did not use any example in $\mathfrak{s l}(V)$, I know it a map from $V$ to $V$, but how it equals $\mathfrak{s l}(n))$,
we have to compute the trace of this map, but please, if yes, do you have an example of it?

thanks in advance,

I only mentioned $\mathfrak{sl}(V)$ to say that it is the same as $\mathfrak{sl}(\dim V)=\mathfrak{sl}(n).$ $n=2$ is ok, but all of the above works for any value of $n \geq 1.$

What do you mean by the first direction? We start with an arbitrary matrix $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and consider it modulo $\mathfrak{sl}(2),$ i.e. modulo trace zero matrices. We thus can write
\begin{align*}
\mathfrak{gl}(2)\ni \begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\c&-a\end{pmatrix}+ \begin{pmatrix}\begin{pmatrix}0&0\\0&d+a\end{pmatrix}\end{pmatrix}\in \mathfrak{sl}(2)+\mathbb{K}
\end{align*}
That would identify elements of $\mathbb{K}$ with $\begin{pmatrix}0&0\\0&\lambda \end{pmatrix}.$ This is possible, but it is better to identify the elements of $\mathbb{K}$ with $\begin{pmatrix}\lambda &0\\0&\lambda \end{pmatrix}$ since this is an ideal in $\mathfrak{gl}(2),$ the center.

So I looked for the equation
\begin{align*}
\begin{pmatrix}a&b\\c&d\end{pmatrix}&= \begin{pmatrix}\lambda &0\\0&\lambda \end{pmatrix} +\begin{pmatrix}a'&b'\\c'&-a'\end{pmatrix}
\end{align*}
This means I have $b'=b\, , \,c'=c$ and $a=\lambda +a'\, , \,d=\lambda -a'.$ Its solutions are $\lambda =\dfrac{a-d}{2}\, , \,a'=a-\lambda =\dfrac{a+d}{2}$ and thus
\begin{align*}
\begin{pmatrix}a&b\\c&d\end{pmatrix}&= \begin{pmatrix}\dfrac{a-d}{2} &0\\0&\dfrac{a-d}{2} \end{pmatrix} +\begin{pmatrix}a-\dfrac{a-d}{2}&b\\c&-a+\dfrac{a-d}{2}\end{pmatrix}\\
&=\underbrace{\begin{pmatrix}\dfrac{a-d}{2} &0\\0&\dfrac{a-d}{2} \end{pmatrix}}_{\in Z(\mathfrak{gl}(2))}+\underbrace{\begin{pmatrix}\dfrac{a+d}{2}&c\\b&-\dfrac{a+d}{2}\end{pmatrix}}_{\in [\mathfrak{gl}(2),\mathfrak{gl}(2)]=\mathfrak{sl}(2)}
\end{align*}

 

[HDB1]

Thank you so much, @fresh_42 ,

I mean, when we want to talk about $\operatorname{sl}(V)$, as a subalgebra of $\text { End } V$,
then how we will know the basis? if we do not want to use matrices., and the dimension of $V$ equals 2.

I do not know whether I write my question properly, or not.

Thanks a lot,

 

[fresh_42]

Thank you so much, @fresh_42 ,

I mean, when we want to talk about $\operatorname{sl}(V)$, as a subalgebra of $\text { End } V$,
then how we will know the basis? if we do not want to use matrices., and the dimension of $V$ equals 2.

I do not know whether I write my question properly, or not.

Thanks a lot,

Just for clarification. If $\dim V=n$ and $\mathbb{M}$ are the matrix spaces then
\begin{align*}
\mathfrak{gl}(V)&=\mathfrak{gl}(n)=\operatorname{End}(V)=\mathbb{M}(n,\mathbb{K})=\mathbb{M}(n)\\
\mathfrak{sl}(V)&=\mathfrak{sl}(n)=[\mathfrak{gl}(n),\mathfrak{gl}(n)]=\{X\in \mathfrak{gl}(n)\,|\,\operatorname{trace}(X)=0\}
\end{align*}
If you do not care about matrices, then simple work with elements $\alpha E+\beta H+\gamma F$ and the multiplications $[H,E]=-[E,H]=2E\, , \,[H,F]=-[F,H]=-2F\, , \,[E,F]=-[F,E]=H.$ This is all you need for the (only) three-dimensional, simple Lie algebra, the $\mathfrak{sl}(2).$

We commonly use as standard matrices
$$
E=\begin{pmatrix}0&1\\0&0\end{pmatrix}\, , \,F=\begin{pmatrix}0&0\\1&0\end{pmatrix}\, , \,H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}
$$

The $\mathfrak{sl}(V)$ and all other classical Lie algebras are more or less defined as matrix algebras. Once you have a basis and the multiplication table, you don't need matrices anymore, but it was matrices where all those came from.

 

[HDB1]

Thank you thank you, @fresh_42

 

[HDB1]

A matrix from $\mathfrak{gl}(V)=\mathfrak{gl}(n)$ modulo its derived algebra $[\mathfrak{gl}(n)),\mathfrak{gl}(n))]=\mathfrak{sl}(V)=\mathfrak{sl}(n))$ is in $\mathbb{K}\cdot \mathbb{1}\cong \mathbb{K}.$

Please, @fresh_42 , bear with me, here what will be the equivalncve classes of $\mathfrak{g l}(2) / \mathfrak{s l}(2) \cong \mathbb{K}$?
I mean as in ${0+3Z,1+3Z, 2+3Z}$ in $\mathbb{Z}_3$.

Thank in advance,

 

[fresh_42]

Please, @fresh_42 , bear with me, here what will be the equivalncve classes of $\mathfrak{g l}(2) / \mathfrak{s l}(2) \cong \mathbb{K}$?
I mean as in ${0+3Z,1+3Z, 2+3Z}$ in $\mathbb{Z}_3$.

Thank in advance,

It will be $\begin{pmatrix}0&0\\0&0\end{pmatrix}+\mathfrak{sl}(2)=\mathfrak{sl}(2)$ and $\begin{pmatrix}1&0\\0&1\end{pmatrix}\cdot \mathbb{K}+\mathfrak{sl(2)}.$ The field $\mathbb{K}$ means that we have all multiples of the identity in the representative of the non-zero equivalence class because - other than with numbers - we have entire vector spaces here.

The quotient is one-dimensional and therefore an abelian Lie algebra:
$$
\left[\mathfrak{gl}(2)/\mathfrak{sl}(2)\, , \,\mathfrak{gl}(2)/\mathfrak{sl}(2)\right]=[\mathbb{K}\, , \,\mathbb{K}]=\{0\}
$$
Edit: This can also be written as
$$
\left[\mathfrak{gl}(2)/\mathfrak{sl}(2)\, , \,\mathfrak{gl}(2)/\mathfrak{sl}(2)\right]=\left[\mathfrak{gl}(2)\, , \,\mathfrak{gl}(2)\right]/ \mathfrak{sl}(2)= [\mathfrak{sl}(2)\, , \,\mathfrak{sl}(2)]/\mathfrak{sl}(2)=\mathfrak{sl}(2)/\mathfrak{sl}(2)=\{0\}
$$

 

[HDB1]

Thank you so much @fresh_42