Understand Witten's "QFT and Jones Polynomials" - how does he get to the twisted Dirac operator L_{-}? Hi,
this is my first post here, so I hope I am asking the question the right way.
I am trying to understand to following piece of algebra:
In his paper, Witten claims that $\int_M Tr(B \wedge DB) + \int_M Tr(\phi \wedge \ast D \ast B) = \langle B , \ast DB \rangle + \langle \phi, D \ast B \rangle$ (where B is a Lie-algebra valued 1-form, $\phi$ is a Lie-algebra valued 3-form, $\ast$ is the Hodge star, D is the covariant derivative with respect to some flat connection, and $M$ is a compact closed Riemannian 3-manifold) can be regarded as a product of the form $\langle H, L_- H \rangle$, where $H = B+\phi \in \Omega^1(M,\mathfrak{g}) \oplus \Omega^3(M,\mathfrak{g})$ and $L_- = \ast D + D \ast$ is what he calls the twisted Dirac Operator acting on 1 and 3 forms. The scalar product just comes from extending the inner products of 1- and 3-forms orthogonally onto the direct sum, i.e. 1-forms and 3-forms are orthogonal w.r.t. this inner product.
Witten does not bother to go into any detail explaining that, so I looked it up in another book, "Differential Topology and Quantum Field Theory" by Charles Nash. Now he claims the following (essentially equation 12.104):
$\langle H, L_- H\rangle = \langle B + \phi, (\ast D + D \ast) (B+\phi) \rangle = \langle B+\phi, \ast D B + D \ast B + D \ast \phi\rangle $ (the other term with $\phi$ drops out because $\phi$ is a 3-form, so $D\phi=0$) $= \langle B ,\ast D B\rangle + \langle B, D\ast \phi\rangle + \langle \phi, D \ast B \rangle$. So far so good, it's the linearity of the inner product and the fact that 1- and 3-forms are orthogonal to each other. Now he continues
\begin{eqnarray}
\langle H, L_- H \rangle = \int_M Tr (B \wedge DB) + \int_M Tr(B \wedge \ast D \ast \phi) + \int_M Tr(\phi \wedge \ast D \ast B)
\end{eqnarray}
\begin{eqnarray}
= \int_M Tr(B\wedge DB) + 2 \int_M Tr(\phi \wedge D^\dagger B)
\end{eqnarray}
where $D^\dagger$ is the codifferential of $D$, i.e. $\langle \alpha, D\beta \rangle = \langle D^\dagger \alpha, \beta \rangle$ for differential forms $\alpha, \beta$ with the right degree.
Now I do not see at all how he gets to the last expression. I don't mind the factor of 2, but I don't see how he manages to get the codifferential in this way. I have tried using Stokes as well as the definition of the codifferential and my calculations say that the last two terms in the first line should cancel. However I have to admit that I did not bother about the Lie-algebra part of the forms, i.e. I basically did it for the abelian case. But I was assured that it shouldn't matter. But apparently, it does...
I am pretty desperate to understand this part, so I would be happy about any kind of help you guys can offer me!
 A: The $L^2$ inner product of $su(2)$ valued $p$-forms on a closed manifold 
$M$ is defined by
$\langle a, b\rangle = -\int_M Tr(a\wedge *b).$
Together with the fact that $*^2=\pm 1$ and taking care with signs, this immediately explains the formula of Witten (just insert a $*^2$ before $DB$ in the first integral).  
To see why the definition is correct, note that $-Tr$ is a positive definite inner product on the lie algebra.  $Tr$ is acting on the coefficients and $*$ is acting on the forms.   The usual $L^2$ inner product on (real or complex valued) $p$-forms is
$\langle a, b\rangle = \int_M a\wedge *b.$ 
Also, the adjoint of $D$ is $*D*$ (up so some sign).
For details, look at the chapter on Hodge theory in Warner's book for ordinary forms; passing to vector-bundle valued forms just requires an inner product on the bundle.
A: Hi moep,
$\left<H,L_- H\right> = \left<B,*DB\right> + \left<B,D*\phi\right>+\left<\phi,D*B\right>$
$= \left<B,*DB\right> + \left<D^* B,*\phi\right>+\left<\phi,D*B\right>$
Now, in Euclidean space, $** =(-1)^{n(D-n)}$, where $D$ is the space dimension and $n$ is the degree of the form. For $D=n=3$, it yields $** = 1$. Thus, the last term changes,
$\left<H,L_- H\right>= \left<B,*DB\right> + \left<D ^*B,*\phi\right>+\left<\phi,**D*B\right>,$
However, in Euclidean space, $D^* = (-1)^{D.n +D+1}*D*$, therefore, $D=3$ and $n=1$ yields, $D^* B= -*D*B$, i.e.,
$\left<H,L_- H\right>= \left<B,*DB\right> + \left<D^*B,*\phi\right>-\left<\phi,*D^*B\right>,$
which gives the result you where pointing out!!!
XD
If I tried with Lorentzian signature the result holds... Can someone point out what are we doing wrong?
P.D.: Sign conventions from Nakahara's book (section 7.9).
A: Well well well, it seems like Witten has played a pretty nasty joke on all of us... And all the authors who copied from him apparently fell for it as well!
But I found a possible resolution to the problem above in "Computer Calculation of Witten's 3-Manifold Invariant" by Freed and Gompf (Commun. Math. Phys. 141, 79-117 (1991)):
The formula (1.27) defines a self-adjoint operator $ (-1)^p (* D + D * )$ acting on 2p+1-forms. They go on talking about the $ \eta $-invariant of this operator, which is precisely what also Witten does in his paper later on.
If we take the operator $L_-$ to be this one, then everything works out perfectly.
So I hope this is the answer. If any expert on this matter could confirm this I would be glad. Otherwise I think this question should be answered, but still I welcome any further comments!
A: I think that there is no problem. Consider your two lines formula.
Let me call 2A the second term of the second line.
The third term of the first line equals A : express the codifferential in term of D and 
of the Hodge star using Stokes.
The second term of the first line equals A : use the definition of the codifferential.
For the two calculus, use the fact that < *a , *b> = < a , b > et * * = identity if one acts on
zero or three forms.
If this indications are not sufficient, I will edit to provide more details.
