# 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!

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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.

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First of all, I think the minus sign in front of the trace is just an overall sign in front of all integrals that doesn't affect the problem at hand. Furthermore, it is just a matter of definition of the Lie-algebra inner product, so -Tr or Tr is merely a symbol. Maybe I should clearify the problem by an explicit calculation. Let's look at the second term: $\int_M Tr(B\wedge *D*\phi) = (-)\langle B, D*\phi\rangle =(-)\langle D*\phi,B\rangle = \int_M Tr(D*\phi \wedge *B)$ Now both $*\phi$ and $*B$ are even forms, so using product rule and Stokes, one would get $-\int Tr(*\phi \wedge D*B)$ – moep Aug 8 '12 at 12:54
To continue: This then is $-(-)\langle *\phi,*D*B \rangle = -(-)\langle *D*B,*\phi\rangle = -\int_M Tr(*D*B \wedge \phi) = -\int_M Tr(\phi \wedge *D*B)$, as $*D*B$ is a 0-from and therefore commutes under the wedge product. The minus sign in bracket just incorporates the possibility of defining the inner product with an minus sign in front of the integral without affecting the result. I don't know where I went wrong, but clearly this minus sign at the end shouldn't be! So please, please point out any thing that has the potential to be wrong! – moep Aug 8 '12 at 12:59
The correct signs and this calculation can be found in many books, including Warner's. They are a bit of a pain to work out, but can all be derived from $a \wedge *b= \langle a, b\rangle d~vol$ and the product rule. – Paul Aug 9 '12 at 14:40

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.

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Thanks for the answer! Unfortunately, I have already tried this route. I found (on Wikipedia and also in several different books on differential gemeometry) that the codifferential has a non-trivial sign depending on whether it acts on even or odd forms, i.e. D^dagger B = (-1)^(degree(B)) D B on a 3-manifold (where *=1). Since in this case B is a 1-form, *DB is (according to this formula) -D^dagger B. You would also get this minus sign if you used Stokes theorem. There it comes from the product rule applied to D(*\phi \wedge *B), where now *\phi is even. – moep Aug 7 '12 at 18:49

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!

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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).

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