Question on the dimension of Killing-form

[gda]

Hi guys! I am getting some sort of contradiction using the definition of the killing-form.
The killing form as a matrix (sometimes called metric) in some basis can be written as:

$$\eta_{ab}=f_{ac}^df_{bd}^c$$

where [ itex ]
f_{ab}^c [ /itex ]
are the structure constants of the Lie algebra. Of course, a,b,c=1...dim(G), where dim(G) is the total number of the elements of the basis of the Lie algebra (the number of independent generators).
So, [ itex ]
\eta [ /itex ] is a [ itex ] dim(G)\times dim(G) [ /itex ]
matrix.

What it confuses me is that, for example, in SO(3,1) the metric tensor is [ itex ]
\eta_{ab}=diag(-1,+1,+1,+1) [ /itex ]
. It has dimension: $4\times 4$ . But if I use the definition above of the Killing-form, I get a matrix with dimension 6\times 6 , because the SO(3,1) has 6 generators..

What I am doing wrong ? or there's some concept wrong, i don't know.

 

[gda]

the latex mode doesn't work?

 

[morphism]

The Killing form and the metric tensor are completely different things. The tensor is something preserved by elements of the Lorentz group - it's a metric on R^4. The Killing form is something defined on the Lorentz group - it's a bilinear form on so(3,1) (which is R^6 as a vector space).

P.S. To close a tex tag use /tex and not \tex.

 

[gda]

ok, thanks morphism. So, a priori, the signature of the tensor metric is different from the signature of the Killing form ? they are different matrix just because of their dimensions.

pd: I've already changed it /--->\ but doesn't seem to work

 

[morphism]

The metric tensor and the Killing form are completely different things. So yes, their signatures aren't related in general.

As for your tex problem, you just need to get rid of the spaces inside the []'s:

[ tex ] should be [tex]

 

[jambaugh]

As mentioned they are different metrics, however...
The action of SO(3,1) on any of its representation spaces will preserve a bilinear form i.e. a metric on that space. For the vector representation that will be the usual Minkowski metric. But the group also acts adjointly on its own Lie algebra ($X\to gXg^{-1}$) and the Killing form is the invariant metric on that Lie algebra.

Further since you can express Lie algebra elements as antisymmetric 2-tensors, when you look at the extension of the Minkowski metric on the space of 2 tensors and project onto the subspace of anti-symmetric 2 tensors the metric you'll get will be ... the Killing form! (or some multiple of it.?).

So for example... using a space-time basis of : $\hat{h},\hat{i},\hat{j},\hat{k}$ with $-\hat{h}\cdot\hat{h}= \hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j} =\hat{k}\cdot \hat{k} = 1$,
You may define the basis of generators of SO(3,1) by taking anti-symmetric tensor products of this basis (acting on vectors by dotting with the metric ...)
The boost generator in the z direction is:
$$\beta_z=\hat{h}\wedge\hat{k} = \hat{h}\hat{k}-\hat{k}\hat{h}$$
Note $\beta_z \cdot \hat{k} =\hat{h}(\hat{k}\cdot\hat{k}-\hat{k}(\hat{h}\cdot\hat{k})= \hat{h}-0=\hat{h}$ likewise $\beta_z\cdot\hat{h} = \hat{k}$.

Now to get the metric on the adjoint representation just double-dot ( $a b : c d = a(b\cdot c) \cdot d = (b\cdot c) (a\cdot d)$). This extends the vector metric onto 2-tensors. For higher rank tensors triple dot, quadruple dot, etc.

$\beta_z : \beta_z = (\hat{h}\hat{k}-\hat{k}\hat{h})\cdot\cdot (\hat{h}\hat{k}-\hat{k}\hat{h}) =\hat{h}(\hat{k}\cdot\hat{h})\cdot\hat{k}- ... =0-(-1)-(-1)+0=2$
this will be the Killing form (maybe a multiple of it but I think exactly it).

If you'd rather work in indexed notation:
The basis for the SO(3,1) Lie algebra being $\omega_{[\mu\nu]} = \mathbf{e}_\mu \wedge \mathbf{e}_\nu$
The Killing form will manifest as:
$$\langle \omega_{[\mu\nu]} , \omega_{[\alpha\beta]} \rangle = K_{[\mu\nu][\alpha\beta]} = g_{\mu\beta}g_{\nu\alpha}$$
Note the two indices of the Killing form $K_{ab}$ are each identified by iterating over anti-symmetric pairs of vector indices:$a \in\{ [12],[13],[14],[23],[24],[34]\}$
[edit: or you can just number them 1 through 6 but then you'll need to map back to the anti-symmetric pairs to express in terms of the vector metric $g_{\mu\nu}$.]

 

[HallsofIvy]

the latex mode doesn't work?

It doesn't if you put spaces in [ tex ] and [ /tex ]!
[ tex ]\int_{-\infty}^\infty e^{-x^2}dx[ /tex ]
without the spaces:
$$\int_{-\infty}^\infty e^{-x^2}dx$$
(Sometimes you need to click on the "refresh" button.)

 

[gda]

ok, thank you all!