My question is what is the relation between N=2 super Yang-Mills and its twisted version topological field theory? After twisting N=2 super Yang-Mills, i.e. diagonally embedding $SU(2)'_R$ into $SU(2)_R \times SU(2)_I$, we get a topological field theory. My question is since N=2 SYM and TQFT are different i.e. one is physical and the other is topological. Why can we use TQFT to calculate partition of N=2 SYM? What are the same for these two different theories?

Update: From the second paper of Trimok, the authors claim that SYM under twist are just redefination. How to understand it?

  • 1
    $\begingroup$ Not a expert, but they are certainly not the same. Following the original paper ($2.14 \to 2.18$ ) or this one ($2.3$,$2.4$,$2.30$), the first thing to do is to decompose supersymmetric charges and fields under the representations of the new twisted symmetry group $SU(2)_L \times SU(2)'_R \times U(1)_R$, and to write some action. The $N=2$ twisted action is then shown to be equivalent to a topological Yang-Mills action (+ extra-terms) $\endgroup$ – Trimok Oct 14 '13 at 10:03
  • 1
    $\begingroup$ The main difference is that the observables in the topological field theory are only a subset of the observables in the physical theory. The subset consists of $Q$-closed operators, where $Q$ is the scalar supersymmetry charge present in the twisted theory. $\endgroup$ – suresh Feb 2 '14 at 1:13

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.