A About universal enveloping algebra

[HDB1]

Please, I have a question about universal enveloping algebra: Let $U=U(\mathfrak{g})$ be the quotient of the free associative algebra $\mathcal{F}$ with generators $\left\{a_i: i \in I\right\}$ by the ideal $\mathcal{I}$ generated by all elements of the form $a_i a_j-a_j a_i-\sum_{k \in I} c_{i, j}^k a_k$. The associative algebra $U(\mathfrak{g}):=\mathcal{F} / \mathcal{I}$ is called the universal enveloping algebra.

1- Please, could I know what is free algebra here, and why universal enveloping algebra is associative?

2- and please, if you could explain to me the universal enveloping algebra of lie algebra $\mathfrak{s l}_2$,Thanks in advance,

 

[HDB1]

Dear, @fresh_42 , I am so sorry, Please, bear with me,

 

[fresh_42]

Please, I have a question about universal enveloping algebra: Let $U=U(\mathfrak{g})$ be the quotient of the free associative algebra $\mathcal{F}$ with generators $\left\{a_i: i \in I\right\}$ by the ideal $\mathcal{I}$ generated by all elements of the form $a_i a_j-a_j a_i-\sum_{k \in I} c_{i, j}^k a_k$. The associative algebra $U(\mathfrak{g}):=\mathcal{F} / \mathcal{I}$ is called the universal enveloping algebra.

1- Please, could I know what is free algebra here, and why universal enveloping algebra is associative?

2- and please, if you could explain to me the universal enveloping algebra of lie algebra $\mathfrak{s l}_2$,Thanks in advance,

I hate the universal enveloping algebra. It is sometimes necessary but in general of little help. It is associative because we construct it to be associative. We gather all we need as elements, which are all elements of the Lie algebra, make an associative algebra out of it, and in order that it still has anything to do with our original Lie algebra, we identify the multiplication of the Lie algebra as if it came from ordinary matrix multiplication.

Example: $\mathfrak{g}=\mathfrak{sl}(2).$

This means we need primarily the elements $\{E,H,F\}.$ But now, consider them as just letters; an alphabet with $3$ letters if you like. Our free algebra now will consist of all linear combinations of any finite words over that alphabet. E.g.
$$
FHE \in \mathcal{F}\; , \; H^3EH\in \mathcal{F}\; , \;FEHEFHEHEFHEF\in \mathcal{F}\; , \;E^2\in \mathcal{F}
$$
We require associativity per construction. And we have the linear span of these words. That thing is huge!

In the second step, we consider the elements $HE-EH-2E\, , \,HF-FH+2F\, , \,EF-FE-H$ which are already in $\mathcal{F}.$ They span a subspace and generate an ideal in $\mathcal{I}\subseteq \mathcal{F}.$ And I have no imagination of what it looks like. The good news is, we do not have to care. We are interested in the quotient (associative) algebra $U(\mathfrak{g})=\mathcal{F}/\mathcal{I}.$ This means the ideal becomes our new zero. But what does it mean if the elements of $\mathcal{I}$ are considered to be zero? It means that
\begin{align*}
HE-EH-2E=0 &\Longleftrightarrow HE-EH=2E\\
HF-FH+2F=0 &\Longleftrightarrow HF-FH=-2F\\
EF-FE-H=0&\Longleftrightarrow EF-FE=H
\end{align*}
Now that is the reason for all the trouble: we have equations in an associative algebra that pretend to be Lie multiplications of matrices. Remember that $[H,E]=HE-EH=2E\, , \,[H,F]=HF-FH=-2F\, , \,[E,F]=EF-FE=H$ as $2\times 2$ matrices with vanishing trace. However, we constructed something like an associative hull of our Lie algebra, therefore the name enveloping. And universal because we put in as many elements as ever possible. Ok, the correct mathematical reason why it is called universal is a bit more sophisticated than that, but not of interest here. Here is the full dose (to deter):
https://ncatlab.org/nlab/show/universal+enveloping+algebra

1.) The equations we get from the quotient by $\mathcal{I}$ allow replacing $HE$ by $2E+EH$ so we can sort $H$ and $E$ and similar for $H,F$ and $E,F.$ Hence the words in $\mathcal{F}$ can be written as linear combinations of $E^pH^qF^r$ with $r,p,q\in \mathbb{N}_0$ which are thus the typical elements in $U(\mathfrak{g}).$ However, this might be different in the cases where the multiplications in $\mathfrak{g}$ are more complicated.

2.) Just a remark about quotient (Lie) algebras. Say we consider the example $\mathbb{Z} / 3\mathbb{Z}$ again. Some authors say factor (Lie) algebras instead of 'quotient' and that $3\mathbb{Z}$ is factored out. Both are crutches since it is neither a quotient nor a factor. However, $\mathbb{Z}/3\mathbb{Z}$ is a partition of $\mathbb{Z}$ into the three sets $0+3\mathbb{Z}\, , \,1+3\mathbb{Z}\, , \,2+3\mathbb{Z}$ which become the new elements. Hence, we have built a quotient along $3\mathbb{Z}$ but we have also factored $\mathbb{Z}$ into the remainder classes $0,1,2$ which we get from a division by $3$. So whenever someone says quotient space or factor space, then a construction $A/B$ is meant.

Don't spend too much thought on it. But keep the construction principle in mind, it occurs on many more occasions, e.g. exterior product spaces (Graßmann algebras; wedge products).

 

[HDB1]

Hence the words in $\mathcal{F}$ can be written as linear combinations of $E^pH^qF^r$ with $r,p,q\in \mathbb{N}_0$ which are thus the typical elements in $U(\mathfrak{g}).$

Thank you so much, @fresh_42 ,

1- please, how we can prove is is linearly independent?2- Please, in this: in semidirect product of lie algebra $\mathfrak{a}:=\mathfrak{s l}_2 \ltimes V_2$,
The Lie algebra $\mathfrak{a}$ admits the basis $\{H, E, F, X, Y\}$ and the Lie bracket is defined as follows
$$
\begin{aligned}
& {[H, E]=2 E, \quad[H, F]=-2 F, \quad[E, F]=H, \quad[E, X]=0, \quad[E, Y]=X,} \\
& {[F, X]=Y, \quad[F, Y]=0, \quad[H, X]=X, \quad[H, Y]=-Y, \quad[X, Y]=0 .} \\
&
\end{aligned}
$$
Let $A=U(\mathfrak{a})$ be the enveloping algebra of the Lie algebra $\mathfrak{a}$.

please why we write the universal enveloping algebra in this way: $A =U\left(\mathfrak{s l}_2\right) \otimes U\left(V_2\right)$
why we don't use the direct sum as in lie algebra $\mathfrak{a} =\mathfrak{s l}_2 \oplus V_2$

Thanks in advance,

 

[fresh_42]

Thank you so much, @fresh_42 ,

1- please, how we can prove is is linearly independent?

That what is linearly independent? The $\mathcal{F}$ in free algebra makes sure that all words are linearly independent as long as they are different. Free means: no relations, no equations, except the associative law and the distributive law, which are necessary to make it an associative algebra. The elements are per construction linearly independent. But things change for $U(\mathfrak{g})=\mathcal{F}/\mathcal{I}$ where we artificially add linear dependencies via $\mathcal{I}$ because we identify e.g. $HE=EH+2E.$

So $\{EH,HE\}$ are linearly independent in $\mathcal{F}$, $\{EH,HE,E\}\subseteq \mathcal{F},$ too.
So $\{EH,HE\}$ are linearly independent in $U(\mathfrak{g})$, $\{EH,HE,E\}\subseteq U(\mathfrak{g})$ is linearly dependent since $1\cdot HE -1\cdot EH -2\cdot E=0$ is a non-trivial linear combination of the zero vector in $U(\mathfrak{g}).$

Note that linear independence in $\mathcal{F}$ can turn into linear dependence in $\mathcal{F}/\mathcal{I}$ because we consider every word in $\mathcal{I}$ as zero.

 

[fresh_42]

2- Please, in this: in semidirect product of lie algebra $\mathfrak{a}:=\mathfrak{s l}_2 \ltimes V_2$,
The Lie algebra $\mathfrak{a}$ admits the basis $\{H, E, F, X, Y\}$ and the Lie bracket is defined as follows
$$
\begin{aligned}
& {[H, E]=2 E, \quad[H, F]=-2 F, \quad[E, F]=H, \quad[E, X]=0, \quad[E, Y]=X,} \\
& {[F, X]=Y, \quad[F, Y]=0, \quad[H, X]=X, \quad[H, Y]=-Y, \quad[X, Y]=0 .} \\
&
\end{aligned}
$$
Let $A=U(\mathfrak{a})$ be the enveloping algebra of the Lie algebra $\mathfrak{a}$.

please why we write the universal enveloping algebra in this way: $A =U\left(\mathfrak{s l}_2\right) \otimes U\left(V_2\right)$
why we don't use the direct sum as in lie algebra $\mathfrak{a} =\mathfrak{s l}_2 \oplus V_2$

This is the general difference between a direct sum, i.e. $\oplus$ and a tensor product $\otimes.$

The direct sum is still a sum: the elements of $\mathfrak{a}$ are all of the form
$$
\mathfrak{a} \ni \underbrace{\alpha E+\beta H+\gamma F}_{\in \mathfrak{sl}(2)} +\underbrace{\delta X +\varepsilon Y}_{\in V_2},
$$
i.e. a sum from an element of $\mathfrak{sl}(2)$ and an element from $V_2$.

The tensor product is all possible (linear combinations of) products of elements from $U(\mathfrak{sl}(2))$ and $U(V_2).$ We need all possible words over the alphabet $\{E,H,F,X,Y\}$ and linear combinations thereof to create the free algebra $\mathcal{F}$ now over $\{E,H,F,X,Y\}.$ The linear combination of those words are unsorted in the free algebra.

The question is: why can we write the products as words from $U(\mathfrak{sl}(2))$ with words from $U(V_2)$ in a sorted manner? They do not commute, so why can we sort them? Here come the quotient algebras into play. The universal enveloping algebras are quotients, i.e. they have e.g. the rules $HE=EH+2E$ and $FX=XF+Y.$ This allows us to sort the elements of $U(\mathfrak{sl}(2))$ to the left. If $XF$ occurs, we replace it with $FX-Y.$ And similar with
\begin{align*}
XE&=EX\\
YE&=EY-X\\
XF&=FX-Y\\
YF&=FY\\
XH&=HX-X\\
YH&=HY+Y\\
\end{align*}
These are the rules we get from the factor $\mathcal{I}.$ And therefore we can sort all $E,H,F$ to the left and write
$$
\underbrace{U( \mathfrak{sl}(2) \ltimes V_2 )}_{ \text{words over E,H,F,X,Y} } \cong \underbrace{ U(\mathfrak{sl}(2))}_{ \text{words over E,H,F} } \otimes \underbrace{ U(V_2)}_{ \text{words over X,Y} }
$$
as sorted words with those rules. Note that $U(\mathfrak{sl}(2))$ is not commutative, e.g. $HE\neq EH$ but $U(V_2)$ is commutative because $XY=YX.$ So we can sort them, but we still have words built from five letters and their linear combinations. So we have products, not only sums.

Maybe you are interested to learn a bit about tensors:
https://www.physicsforums.com/insights/what-is-a-tensor/

 

[HDB1]

Thank you so so much, @fresh_42, you are great ,

Please, could you explain about associative operation in universal enveloping algebra, I need example, please:

Do you mean: $F(H^2EX) = (FH^2E)X= FH^2(EX)= FH^2(EX)$

Thanks in advance,

 

[fresh_42]

Thank you so so much, @fresh_42, you are great ,

Please, could you explain about associative operation in universal enveloping algebra, I need example, please:

Do you mean: $F(H^2EX) = (FH^2E)X= FH^2(EX)= FH^2(EX)$

Thanks in advance,

Yes.

Associativity says that we do not have to bother the order of multiplications. It does not mean that we can exchange elements, so $EF\neq FE,$ but $E(FH)=(EF)H.$ That's why we are allowed to write simply $EFH$ since the order of multiplications doesn't play a role. However, the order of letters does!

E.g. Lie multiplication $4E=[H,[H,E]] \neq [[H,H],E]=0$ is not associative. But in $\mathcal{F}$ or $U(\mathfrak{g})$ we have associativity and $F(H^2EX)=F(H(H(EX)))=(((FH)H)E)X.$

 

[HDB1]

$1\cdot HE -1\cdot EH -2\cdot E=0$ is a non-trivial linear combination of the zero vector in $U(\mathfrak{g}).$

Note that linear independence in $\mathcal{F}$ can turn into linear dependence in $\mathcal{F}/\mathcal{I}$ because we consider every word in $\mathcal{I}$ as zero.

Please, here: we have to put every element in $U(\mathfrak{g})$ equals zero?
or just put any element as linear combination of $E^p H^q F^r$.

Thanks in advance,

 

[HDB1]

Yes.

Associativity says that we do not have to bother the order of multiplications. It does not mean that we can exchange elements, so $EF\neq FE,$ but $E(FH)=(EF)H.$ That's why we are allowed to write simply $EFH$ since the order of multiplications doesn't play a role. However, the order of letters does!

E.g. Lie multiplication $4E=[H,[H,E]] \neq [[H,H],E]=0$ is not associative. But in $\mathcal{F}$ or $U(\mathfrak{g})$ we have associativity and $F(H^2EX)=F(H(H(EX)))=(((FH)H)E)X.$

Thank you, thank you,

 

[fresh_42]

Please, here: we have to put every element in $U(\mathfrak{g})$ equals zero?
or just put any element as linear combination of $E^p H^q F^r$.

Thanks in advance,

$U(\mathfrak{g})=\mathcal{F}/\mathcal{I}$ which makes all words to zero that are contained in $\mathcal{I}.$ This ideal is generated by the Lie multiplications. E.g. if $[A,B]=C$ then $\mathcal{I}$ contains the vector $AB-BA-C.$ If this is zero, per construction of the quotient, then $BA=AB-C.$ Let us further assume that $[A,C]=0.$ This becomes $AC-CA=0$ in $\mathcal{I}.$

Hence any word in $U(\mathfrak{g})$ that contains $BA$ can be replaced by $AB-C.$ E.g.
\begin{align*}
UWBA^2X&=UW(BA)AX=UW(AB-C)AX=UW(AB)AX-UWCAX \\
&=UWA(BA)X-UWCAX=UWA(AB-C)X-UWCAX\\
&=UWA^2BX-UW(AC)X-UW(CA)X =UWA^2BX-UW(AC)X-UW(AC)X\\
&=UWA^2BX-2\cdot UW(AC)X
\end{align*}
This was an arbitrary example for a virtual enveloping algebra with basis vectors $U,W,X,A,B$ and two (among others) multiplications $[A,B]=C\, , \,[A,C]=0$ in the corresponding Lie algebra. That's how it works, i.e. how words with $BA$ and $CA$ in it can be sorted to have $AB$ and $AC$ instead. I wouldn't spend too much thought on it.

 

[HDB1]

Thank you so much, @fresh_42 , I really appreciate everything you've done,

- I have another question, Please, how we can compute the center of $U(\mathfrak{s l}(2))$?

Thanks in advance,

 

[fresh_42]

Thank you so much, @fresh_42 , I really appreciate everything you've done,

- I have another question, Please, how we can compute the center of $U(\mathfrak{s l}(2))$?

Can we? Well, let's see.

$U(\mathfrak{sl}(2))$ are all linear combinations of words over the alphabet $\{E,H,F\}.$ The center of $U(\mathfrak{sl}(2))$ are all elements $w\U(\mathfrak{sl}(2))$ such that $w\cdot v =v\cdot w$ for all $v\in U(\mathfrak{sl}(2)).$ This means $w\cdot v-v\cdot w=0$ for all $v\in U(\mathfrak{sl}(2)).$ For all $v\inU(\mathfrak{sl}(2))$ means in particular for all $v\in \{E,H,F\}.$ The zero in $U(\mathfrak{sl}(2))=\mathcal{F}/\mathcal{I}$ are all words in the ideal $\mathcal{I}.$ It is generated (as an ideal) by the words $HE-EH-2E\, , \,HF-FH+2F\, , \,EF-FE-H.$

These relations also allow us to write every element $w\in U(\mathfrak{sl}(2))$ as linear combinations of $E^pH^qF^r$ for some $p,q,r \in \mathbb{N}_0.$

Everything until here was for free. Now come the ugly calculations. We know that all words from $\mathcal{I}$ are in the center since they are considered to be zero. The question is: does have $U(\mathfrak{sl}(2))$ an element in its center that is not in $\mathcal{I},$ i.e. different from zero?

My suspicion is that this is not the case and that $Z(U(\mathfrak{sl}(2)))=\{0\}=\mathcal{I}.$ But how to prove it? There are simply so many elements in the enveloping algebra:
$$
E,H,F,EH,EF,HF,EHF,E^2,H^2,F^2,E^2H,E^2F,H^2F,E^3,\ldots
$$
and all linear combinations of them. I have no idea how to tackle that. It is clear that $E,H,F$ are not in the center so maybe we could work with a minimal length. However, the possibility of linear combinations makes it difficult. A general element looks like
$$
w=\sum_{(p,q,r)\in \mathbb{N}_0^3} \lambda_{(p,q,r)}E^pH^qF^r
$$
with only finitely many $\mathbb{K}\ni \lambda_{(p,q,r)}\neq 0,.$

Now investigate $wE-Ew \in \mathcal{I}\, , \,wH-Hw \in \mathcal{I}\, , \,wF-Fw \in \mathcal{I}.$

 

[fresh_42]

By the way; if you want a shortcut through Humphrey's book which is mainly about semisimple Lie algebras because these are which are classified, then you may read:

Basics: https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/
Structures: https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-structures/
Representations: https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/

 

[HDB1]

Thank you so ,much, @fresh_42.

 

[Infrared]

My suspicion is that this is not the case and that $Z(U(\mathfrak{sl}(2)))=\{0\}=\mathcal{I}.$ But how to prove it? There are simply so many elements in the enveloping algebra:

The Casimir element is always central. In this case, it is $\frac{1}{2}H^2+H+2FE$ according to wikipedia.

 

[fresh_42]

The Casimir element is always central. In this case, it is $\frac{1}{2}H^2+H+2FE$ according to wikipedia.

And now that you say it, $1$ should be in the center, too.

 

[HDB1]

And now that you say it, $1$ should be in the center, too.

Thank you so much, @fresh_42 , please, do you know how we get this element?
I mean this form: $\frac{1}{2} H^2+H+2 F E$
I found it in page 28.

Thanks in advance,

 

[fresh_42]

We have the representation as matrixes here:
\begin{align*}
\text{Set }X_1=E\, , \,X_2=H\, , \,X_3=F.&\\
\phi\, : \,L=\operatorname{lin\,span}\{X_1,X_2,X_3\}&\longrightarrow \mathfrak{gl}(2)\\
\alpha E+\beta H+\gamma F&\longmapsto \begin{pmatrix}\gamma &\alpha\\\beta & -\gamma \end{pmatrix}
\end{align*}
Then we need a dual basis $\{Y_1,Y_2,Y_3\}\subseteq L,$ i.e. a basis that satisfies $\beta(\phi(X_i)\, , \,\phi(Y_j))=\operatorname{trace}(\phi(X_i)\, , \,\phi(Y_j))=\delta_{ij-}.$

Humphreys says that such a basis is $\left\{Y_1=F\, , \,Y_2=\dfrac{1}{2}H\, , \,Y_3=E\right\}.$

You should check this. E.g.
\begin{align*}
\operatorname{trace}(\phi(X_1)\, , \,\phi(Y_1))&=\operatorname{trace}(\phi(E),\phi(F))=
\operatorname{trace}\left(\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}\right)=1\\
\operatorname{trace}(\phi(X_1)\, , \,\phi(Y_2))&=\operatorname{trace}\left(\phi(E),\dfrac{1}{2}\phi(H)\right)=
\operatorname{trace}\left(\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}1/2&0\\0&-1/2\end{pmatrix}\right)=0\\
\operatorname{trace}(\phi(X_1)\, , \,\phi(Y_3))&=\operatorname{trace}(\phi(E),\phi(E))=
\operatorname{trace}\left(\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}\right)=0\\
&\text{etc.}
\end{align*}

The Casimir element is defined as
\begin{align*}
c_\phi(K)&=\sum_{k=1}^3\phi(X_k)\phi(Y_k)=\phi(E)\cdot \phi(F)+\phi(H)\cdot\phi((1/2)H)+\phi(F)\cdot\phi(E)\\
&=\begin{pmatrix}0&1\\0&0\end{pmatrix}\cdot \begin{pmatrix}0&0\\1&0\end{pmatrix}+
\begin{pmatrix}1&0\\0&-1\end{pmatrix}\cdot\begin{pmatrix}(1/2)&0\\0&-(1/2)\end{pmatrix}+
\begin{pmatrix}0&0\\1&0\end{pmatrix}\cdot\begin{pmatrix}0&1\\0&0\end{pmatrix}\\
&=\begin{pmatrix}1&0\\0&0\end{pmatrix}+\begin{pmatrix}(1/2)&0\\0&(1/2)\end{pmatrix}+\begin{pmatrix}0&0\\0&1\end{pmatrix}\\
&=\dfrac{3}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}
\end{align*}

Humphreys considers $E\, , \,H\, , \,F$ as matrices, too, so $\phi$ becomes the identity map, e.g.
$$
\phi(F)=\phi\left(\begin{pmatrix}0&0\\1&0\end{pmatrix}\right)=\begin{pmatrix}0&0\\1&0\end{pmatrix}
$$
So Humphreys doesn't need $\phi$ anymore and he can write
$$
c_\phi=c_\phi(\beta)=c_{\operatorname{id}}(\text{trace})=EF+\dfrac{1}{2}H^2+FE=xy+(1/2)h^2+yx
$$

 

[fresh_42]

Note that $EF-FE-H=0$ in $U(\mathfrak{sl}(2)).$ Thus
$$
\dfrac{1}{2}H^2+H+2FE=\dfrac{1}{2}H^2+(EF-FE)+2FE=EF+\dfrac{1}{2}H^2+FE
$$
which explains that @Infrared 's formula in post #16 is the same as Humphreys's formula of $c_\phi$ if we consider the Casimir element as an element in the universal enveloping algebra of $\mathfrak{sl}(2).$