# What is the definition of quotient Lie algebra?

• A
• HDB1
In summary: B) So if we use the commutator bracket (the Lie bracket), what will happen with matrix ##X##?C) And what will happen with ##\lambda \mathbb{1}##?D) So what will happen with ##A##?If you do this for all possible such matrices, you will find that ##[\mathfrak{sl}(2),\mathfrak{sl}(2)]=\mathfrak{sl}(2).## The center is not contained in the commutator, and it is only the multiples of the identity matrix.
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,

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

HDB1 said:
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
fresh_42 said:
\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?

Last edited:
HDB1 said:
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*}

HDB1
HDB1 said:
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
fresh_42 said:
$$\bar A = A+\{\lambda \cdot \mathbb{1}\,|\,\lambda \in \mathbb{K}\}= X$$
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,

HDB1 said:
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,

HDB1 said:
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*}

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,

HDB1 said:
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
fresh_42
fresh_42 said:
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,

HDB1 said:
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
$$\left[\mathfrak{gl}(2)/\mathfrak{sl}(2)\, , \,\mathfrak{gl}(2)/\mathfrak{sl}(2)\right]=\left[\mathfrak{gl}(2)\, , \,\mathfrak{gl}(2)\right]/ \mathfrak{sl}(2)= [\mathfrak{sl}(2)\, , \,\mathfrak{sl}(2)]/\mathfrak{sl}(2)=\mathfrak{sl}(2)/\mathfrak{sl}(2)=\{0\}$$

Last edited:
HDB1

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