# Connection between killing forms and metrics

1. Apr 1, 2009

### Bobhawke

I wasnt quite sure where to put this thread. This question occurred to me while I was looking at the group theory of standard model groups so I thought I'd put it here.

Anyway, here is my question: One can define the Killing form for a group by taking the trace of two generators. One can then choose a basis in which this quantity is diagonal and has entries + or - one. This appears to have the same form as a metric, and in fact in my notes the same symbol is used to denote the metric and the killing form. They both have many similar properties. But it doesnt seem to me like there should be any connection between the two - the killing form is a feature of the group, whereas the metric is a feature of the space we are working in. Are they indeed the same thing, and if they are, then why?

2. Apr 2, 2009

### genneth

On a (pseudo-)Riemannian manifold, a metric is simply a privileged symmetric invertible 2-tensor. If you have some suitably behaved differential manifold, like a Lie group, arbitrarily picking such a tensor will essentially make it into a Riemannian manifold.

3. Apr 2, 2009

### blechman

it shouldn't surprise you too much that the killing form is a metric on the Lie algebra. If you have a tangent space on a manifold, you can define a metric on that tangent space by taking the suitably-defined dot products of all the basis vectors.

The Lie algebra is the tangent space of the Lie group manifold, the generators of the group form a basis for this tangent space, and the killing form is the suitably-defined dot product. So it really is the same story.

4. Apr 2, 2009

### Bobhawke

Ok but does the metric defined on the group space have anything to do with the metric for the vector space on which a matrix representation of the group acts?

5. Apr 3, 2009

### rntsai

It's not very clear what you mean by "metric for the vector space on which ... " but I think the quick answer is no...they have nothing to do with each other. For one the killing form (metric) is over an N_g dimensional space where N_g is the number of infinitesimal generators of the group. Different reps of the group have different dimensions.... so at the very least you're dealing with different dimensional spaces. If you just look at the adjoint rep, then that will be N_g dimensional, but even in this case I'm not sure of there are any deep connections between the killing form and any other metric you migh be considering here....

6. Apr 8, 2009

### Bobhawke

Could someone explain to me why the tangent space to the group manifold at the origin is the lie algebra? I can almost see why - If you differentiate a group element you pull down a generator from the exponential, and then set it equal to the identity means you get just the generator. But you get that from differentiating a group element, to get a tangent space you have to differentiate vectors in the group manifold which is a different thing.

7. Apr 8, 2009

### xepma

I'm not enterily sure what you mean by:

There is no such thing as "vectors in the group manifold". The group manifold is a parameter space, in which the set of parameters have the special property that they form a manifold. And manifolds have the 'special' property that you can define a tangent space. The group elements correspond to elements of the manifolds ("coordinates"). The tangent space near the origin is identified (or defined as) the corresponding Lie algebra.

A tangent space at some point x of a manifold is defined through differentiation of curves (not vectors) through this point x. The Lie algebra corresponds to the tangent space of the point x= identity of the group.

As a corrollary, vector elements of this tangent space act as differential operators on functions on the manifold. It's a bit of a mathematical structure, but if you seen it once you'll start to appriciate it.

8. Apr 8, 2009

### Bobhawke

Yeah ok I meant curves not vectors.

The point is why is it that the tangent space of the group space at the origin is the lie algebra? I mean I guess you can define it to be so, but then my question simply becomes this: why is it that the tangent space of the group manifold at the origin is a space with the group generators as its basis?

9. Apr 9, 2009

### blechman

again, i think it's a matter of definition. a Lie group/manifold is one whose tangent bundle is a Lie algebra. That is: not all manifolds are Lie groups, only the ones for which this is true!

[I'm sure my wording above is very sloppy, but hopefully you get the idea].

But perhaps someone who has thought more about these issues can say more.