I am in trouble with calculation details of Witten's Two dimensional Gauge Theories Revisited. My questions is about (3.21) and (3.27). From section 3, we have $$\delta A_i=i\epsilon \psi_i\\ \delta \psi_i=-\epsilon D_i\phi \\ \delta \phi=0\\ \delta \lambda=i\epsilon\eta\\ \delta\eta=\epsilon[\phi,\lambda]\\ \delta \chi=\epsilon H\\ \delta H=i\epsilon [\phi,\chi]\\ V= \frac{1}{h^2}\int_\Sigma d\mu {\rm{Tr}} (\frac{1}{2}\chi(H-2\star F)+g^{ij}D_i\lambda\psi_j) $$ Then the (3.21) is given by $$ L=-i\{Q,V \}\\= \frac{1}{h^2}\int_\Sigma d\mu {\rm{Tr}}(\frac{1}{2}(H-f)^2-\frac{1}{2}f^2 -i\chi\star D\psi+iD_i\eta\psi^i +D_i\lambda D^i\phi +\frac{i}{2}\chi[\chi,\phi]+i[\psi_i,\lambda]\psi^i ) $$

My questions are followings:

1) What is the relations between $\delta \bullet$ and $\{Q,\bullet\}$? Is there a minus sign when Q-operator crossing $\lambda$ or $\chi$ ? I always found different signs in terms $ -i\chi\star D\psi$ and $ D_i\lambda D^i\phi$ when I use relation $\delta\bullet=-i\epsilon\{Q,\bullet \}$.

2) Why the two terms $ -i\chi\star D\psi$ and $ D_i\lambda D^i\phi$ have different signs in (3.21) and (3.27)? Anything wrong with it?

======================================================================== =====

Based on discussion with Trimok, (Thanks to Trimok but this does not mean Trimok agrees with my views) I think I understand part of the problem. First, the relations is $\delta\bullet=-i\epsilon\{Q,\bullet \}$. $\delta$ pass $\chi$ and $\eta$ will give a minus sign. So in the formula (3.17), some of terms should be $ i\chi\star D\psi-D_i\lambda D^i\phi$ and (3.27) is correct. While I have a new question now about (3.28).

From (3.27) to (3.28), the procedures are the followings: First, perform variation of H, and then integrate out H. Second, perform variation of f, and then integrate out $\lambda$. Second, perform variation of $\eta$, and then integrate out $\chi$. After these calculation, the last term in (3.28), i.e. $\frac{1}{t^2} \frac{1}{2} (-D_kD^k+i[\psi_k,\psi^k])^2 $ does not appear. It seems from variation of $f$ from first part of (3.28). How to produce such a term in integrating out fields?

  • 1
    $\begingroup$ 1) Yes, you have $\delta\bullet=-i\epsilon\{Q,\bullet \}$, so $L = \frac{1}{\epsilon} \delta V$. The sign of the first term ($-i \chi \star D\psi$) is correct. Considering $ \frac{1}{\epsilon} -\delta(\chi \star F)$, with $\delta F = i \epsilon D\psi$, you get one term $-i \chi \star D\psi$. The second term seems incorrect, I would have written $-D_i\lambda D^i\phi$, because $\delta \psi_i=-\epsilon D_i\phi$. 2) Finally, I don't understand the signs in (3.21 and 3.27), I would have written (in 3.21 and 3.27): $-i\chi\star D\psi+iD_i\eta\psi^i -D_i\lambda D^i\phi$ $\endgroup$ – Trimok Jul 8 '13 at 8:55
  • $\begingroup$ Thanks for your explaination. Do we have such relation here $Trg^{ij}D_i\lambda \psi_j=Trg^{ij}\psi_j D_i\lambda $? In that case, there will not be such problem as $Q$ crossing $\lambda$ or$\chi$ gives a minus sign as usual. $\endgroup$ – thone Jul 8 '13 at 9:01
  • $\begingroup$ In that case, Q-operator across $D_i\lambda$ will give a minus sign. So $i\chi\star D\psi$ in the end will be + rather than minus and $D_i\lambda D^i\phi$ will be plus sign? $\endgroup$ – thone Jul 8 '13 at 9:17
  • $\begingroup$ If you agree with $L = \frac{1}{\epsilon} \delta V$, you simply apply $\delta$, that is $\delta(ab) = \delta (a) b + a \delta(b) $ $\endgroup$ – Trimok Jul 8 '13 at 9:20
  • $\begingroup$ I do not think you have $\delta (ab)=(\delta a) b+a(\delta b)$ always. $\endgroup$ – thone Jul 8 '13 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.