Lie algebras:Killing form

[Dogtanian]

OK, firstly I hope this is the rigth place for my question. I'm in a bit of a problem. I need to be able to calucalte the Killing for for a Lie algebra by next wek, but I'm stuck and won't be able to get any help in 'real life' until Friday, not leaving me enough time to sort out my problem. So I was hoping someone here might be able to show me some pointers.

 

 

So I have the Lie algebra of all upper triangular 2x2 matrices and am using the basis

(1 0) (0 1) (0 0)

(0 0),(0 0),(0 1).

(I hope these matrices turn out OK: they should be 3 2x2 matrices.)

 

The Killing form is defined as K(X,Y) = trace(adX adY) for all X,Y in the Lie algebra, where ad is the adjoint, defined on Z by adX(Z) = [X, Z] = XZ-ZX.

 

I know I have to consider adX and adY as linear transformations and work out their matrices (and hence multiply them together and find the trace). What I'm not sure is how to go about this, and at what point do I use the bases instead of a general element from the Lie algebra?

 

Any help would be much appreciated, as I've spent a fair bit of time on this and got no where.

 

I also have to so the same for the special linear Lie algebra sl_2© (C=complex nubers), but I think if I have some hints for the triangular one I should hopefully be able to work the special linear one out my self.

[Dave]

Unfortunately I don't do Lie Algebras until next year. However, I've moved this to Modern/Abstract Algebra, since that's probably the best place for it

[matt grime]

Let the basis be e,f,g in the same order as you wrote them. then what is the linear map defined by ad(x) fpr some x in the algebra?

 

It sends e to [xe]=0, t to [xf] =xf-fx, and g to [xg]=xg-gx=0

 

So, just write out X as

 

a b

0 c

 

and mutliply out the above expressions and write [xe] as a combination of basis vectors. [xe] is simply -b.f. Repeat for where it sends f and g, and you've got how it acts on the basis as a linear map thus you can write out its matrix wrt that basis just like any other map.

[Dogtanian]

Thank you.

 

I actually made a mistake, when I said I needed to do this for the upper triangular 2x2 matrices. I really need to do it for the derived Lie algebra of the 3x3 upper triangular matrices (and hence show the Killing from is zero for all matrices of this derived Lie algebra). But your hints have helped me out at first...

 

Is this right:

 

We have the basis e,f,g=

010 001 000

000 000 001

000 000 000

 

For a general X of the form

0 a b

0 0 c

0 0 0

 

we have

adX(e) = -c.f

adX(f) = 0

adX(g) = a.f

 

Now, what I need to do is to show the Killing form ios zero in all cases of X, Y being matrices being made up from the basis e,f,g.

 

Would I now be able to consider X being each of e,f,g in turn in the above and see the results for adX(e) etc in each case (since the Killing form is bilinear). From this I get

ade(e)= 0

ade(f)= 0

ade(g)= f

 

adf(e)= 0

adf(f)= 0

adf(g)= 0

 

adg(e)= -f

adg(f)= 0

adg(g)= 0

 

What, if any, matrix would I be able to work out for the whoe transformation now? (I think this is where my mind is stuck)

 

Would it be, respectively:

ade, adf, adg

 

000 000 0-10

000 000 0 00

010 000 0 00

?

 

Thus we'd see that the mutliplication of any two of these matrices together would result in all zero matrices, all of whih have zero trace and all of which mean the Killing form between an two pairs of basis elements is zero. Hence by bilinearity we see that for any two matrices in the derived Lie algebra we get the Killing form to be zero, hence we see by Catan's Critrion for solvability, that the Lie algebra of all upper triangular 3x3 matices is solvable.

 

Is this the case?

I'll think about the sl_2© example now...

[matt grime]

In order to show that the killing form is degenerate it suffices to find two elements X and Y such that Tr(Ad(X)Ad(Y)) =0, and in the above it so happens you can pick X and Y any two basis elements - you may not be so lucky in general, but yes you've got the gist. I posted a better reply at physicsforums.com

[Dogtanian]

Here's my go so far at the sl_2© case....are there any mistakes?

 

In calculating the Killing form here, I need to show that sl_2© is semisimple.

So I need to show the Killing form is non-degenerate for sl_2©. Due to bilinearity, all I need to show is if K(X,Y)=0 for all X, then Y=0.

 

I can do this by considering the basis

e,f,g

 

(1 0) (01) (00)

(0-1) (00) (10)

 

and hopefully show K(e,Y)=0, K(f,Y)=0, K(g,Y)=0 implies Y=0.

 

Let's start by calculating adX for each basis element of each basis element.

 

ade(e)= 0

ade(f)= 2f

ade(g)= -2g

 

adf(e)= -2f

adf(f)= 0

adf(g)= -e

 

adg(e)= 2g

adg(f)= -e

adg(g)= 0

 

This represents the matrices for ade, adf, adg as the following 3 3x3 matrices

 

(000) (0 -20) (002)

(020) (0 00) (-100)

(00-2) (-1 00) (000)

 

I can can then work out ad(e_i)ad(e_j) for all possible combinations for e_i, e_j from e,f,g.

 

Now, I think I will get all much matrices to have zero trace(I haven't calculated then all yet), no matter whether Y is zero or not. I don't think this is what I want....have I gone wrong somewhere?

[matt grime]

Of course K(e,x)=K(f,X)=K(g,X)=0 implies X=0, but that isn't what you want to show, since it may be that all three things are nonzero but that K is still degenerate:

 

eg suppose K(e,X)=1, K(f,X)=-1 K(g,X)=1, then K(e+f,X)=0.

 

You need to actually calculate ad(X)ad(Y) where X and Y some general elements of sl_2.

 

I will give you for free (and because it's effing tedious) that

 

K(X,Y) = 4tr(XY)

 

where tr is the trace.

 

In particular if the basis is ordered as f,e,g ie the diagonal one in the middle that the matrix of K is given by:

 

[math]\left( \begin{array}{ccc} 0 & 0 & 4 \\ 0 & 8 & 0 \\ 4 & 0 & 0 \end{array} \right)[/math]

 

the latex isnt' appearing so it's:

 

004

080

400

 

in plain ascii

[Dogtanian]

Thanks for telling me K(X,Y) = 4tr(XY). But I thought I'd try to work it out myself anyway as I need it to show how one gets it.

 

I am also able to work out the matrix rep of K you gave.

 

Now, how do I go from here? Do I use this matrix and apply it to a general Y, set this to zero and show that Y must always be zero in this instance?

 

So, I could write a general Y not in the form of a matrix like

a b

c-a

but as a vector

a

b

c

 

 

as by doing this you clearly see by multiplying the matrix for K with such a vector and setting the answer to zero you must have a=b=c=0.

[matt grime]

You must show that the bilinear form it defines is non-degenerate, which is straight forward. If v is any vector in the space v=(a,b,c) define w=(c,b,a) then

 

K(v,w) = |a|^2+|b|^2+|c|^2

 

You cannot apply this matrix to a general Y since it is a *bilinear form* thus needing two inputs. You must show that if v is in sl_2 that K(v,w)=0 for all w implies v=0, or as above that for any v there is some w with K(v,w)=/=0

[Dogtanian]

So, say we to show v =0 for all w.#

 

Let w=

a b

c-a =

(b,a,c)

 

and v=

x y

z-x =

(y,x,z)

 

Then wKv= 8.a.x + 4.c.y + 4.b.z = 0

 

Now is it simply that for this to be true for all a,b,c then x=y=z=0 automatically?

[matt grime]

The answer to the last question is "yes", and I've proved at least twice in this thread that K is nondegenerate on sl_2, by showing that for any v there is a w such that K(v,w) is not zero, and hence K is nondegenerate.

[Dogtanian]

Oh, I get what you did now. You took the contrapositive of the statement:

 

instead of proving

'if K(v,w)= 0 for a ll w then v=0'

 

you proved

 

'if v=/=0 then there exists a w such that K(v,w)=/=0'.

 

I think I get it now.

 

So to do this question fro the start you find the matrix K=

004

080

400

in the way it was done above.

Then you pick a non-zero v = (a,b,c), say, and fnd a w such that K(v,w)= vKw=/=0. In the case above you found that w=(c,b,a) will work.