# Gauge-invariant measure in LQG

1. Oct 28, 2007

### kakarukeys

Could someone explain to me why we use a gauge-invariant and diffeomorphism-invariant measure on the quantum configuration space? Is it because we want the inner product to be invariant under gauge transformations. What is a gauge-invariant measure anyway?

see
http://arxiv.org/abs/hep-th/9305045

Last edited: Oct 28, 2007
2. Oct 28, 2007

### f-h

A measure invariant under the action of the group. For example the measure on the real line R invariant under the translation group is the usual dx:

[TEX]\int_R dx f(x) = \int_R dx f(x+a)[/TEX]

3. Oct 28, 2007

### f-h

And we use them to create gauge invariant theories with gauge invariant inner products.

4. Oct 28, 2007

### kakarukeys

if there is no gauge-invariant measure available, can a gauge-invariant inner product be constructed?

5. Oct 29, 2007

### BenTheMan

Don;t know much about LQG, but assuming it is constructed along the lines of a normal quantum theory...

gauge invariance tells you how to build your theory. You impose gauge invariance, and that gives you a guide as to how to procede. In order to do a calculation, though, you have to pick a gauge---so you use gauge invariance as a tool to write down a langrangian, or something, then you destroy gauge invariance to do calculations. So in a sense gauge invariance is not physical.

Without the gauge invariance, you could still construct a lagrangian. No problem at all---it is still possible to build a theory and write down a lagrangian. This lagrangian will correspond to something that you would have gotten after you chose a gauge in the previous problem.

The diffeomorphism invariance is just the statement that the answer shouldn't depend on the coordinates you use to describe it. It is another type of gauge invariance, in a sense. You write down a lagrangian, given that you have diffeomorphism invarinace. Then you choose a set of coordinates to do calculations.

In answer to your second question, no. But you CAN construct inner products.

6. Oct 29, 2007

### kakarukeys

Is it right to say? Being able to find a gauge-invariant measure in loop quantization is a big achievement because if we performed a traditional canonical quantization we would be using the ill-defined measure $$dA^i_a$$, inner product: $$\int\Phi^*[A^i_a]\Psi[A^i_a]dA^i_a$$
They are not gauge-invariant.

($$A^i_a\tau_i\otimes dx^a$$: SU(2) connection 1-form)

Last edited: Oct 29, 2007
7. Oct 29, 2007

### BenTheMan

what is the measure A over? A characterizes some gauge manifold, right, so you pick a Fadeev-Popov delta function or something?

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