Action of the Gauge Group on the Configuration Space of Seiberg-Witten Theory

Question

In Seiberg-Witten theory, the group of gauge transformations is $Map(M,S^1)$.

For a configuration $(A,\psi)$, where $A$ is a unitary connection on the determinant line bundle, $\psi$ is a spinor, and $g:M\to S^1$ a change of gauge, we have

$(g, (A,\psi)) = (gAg^{-1} - g^{-1}dg, g\psi)$.

Where do we get this action from, and why is this the correct action to look at?

P.S. Remarks on notation: $g^{-1}$ is not the inverse map, but the map $x\mapsto g(x)^{-1}$. Also people usually write just $A - g^{-1}dg$ since $U(1)$ is abelian but I'm writing it with $gAg^{-1}$ to make clear I know what the expression means, I just wanna know how we derive it.

Answer 1

If you are asking for a physicists point of view, the argumentation is usually as follows:

So you want to formulate an action for sections $\psi$ in some vector bundle, say you write something like $$ S[\psi]=\int\overline{\psi}~i\gamma^\mu\partial_\mu\psi $$ in physicists notation (local coordinates). So formally, this is invariant if you replace $\psi\mapsto\text{e}^{i\alpha}\psi$ with $\alpha\in\mathbb{R}$, or similarly, other "constant gauge transformations" depending on where $\psi$ should take values. Then physicists say (e.g. in an undergrad QFT lecture), that "for some relativistic reasons" any global gauge symmetry must be a local one, formally, that $\alpha$ should be allowed to be a function on spacetime. Then the above action is not invariant anymore, but replacing it by something like $$ S[\psi,A]=\int\overline{\psi}~i\gamma^\mu(\partial_\mu+iA_\mu)\psi+\mathcal{L}_\text{free}(A_\mu) $$ with another field $A_\mu$. This is invariant under $\psi\mapsto\text{e}^{i\alpha}\psi$ with $\alpha:M\to\mathbb{R}$ if you additionally claim that $A$ transforms as $A_\mu\mapsto A_\mu\pm \partial_\mu\alpha$ (i don't remember the sign). Hereby, $\partial_\mu+iA_\mu$ is called covariant derivative.

If $\psi$ is a section in some vector bundle, $g\psi$ is just the gauge transformed section, where simultaneously the connection form part of the configuration transforms as connection forms usually do, and your functional depending on $\psi$ and $A$ should be invariant under simultaneously transforming them both.

From a more mathematical point of view, you need to choose a connection form/a gauge in order to take derivatives in a vector bundle and the physics should not depend on this choice. Moreover, in a vector bundle two covariant derivatives are connected via a vector bundle automorphism/gauge transformation. These automorphisms can be identified with elements of your group $\text{Map}(M,G)$, acting as $\psi\mapsto g\psi$. On the other hand, this vector bundle automorphism can be associated with a principal bundle automorphism of the frame bundle (aka gauge transformation), and this gauge transformation acts on local connection forms as $A\mapsto gAg^{-1}-g^{-1}\text{d}g$. So on a configuration $(A,\psi)$, a gauge transformation acts by simultaneously transforming $\psi\mapsto g\psi$ and $A\mapsto gAg^{-1}-g^{-1}\text{d}g$ and invariance under this action guarantees that the physics does not depend on the choice of covariant derivative.

So you said "I just wanna know how we derive it", I don't know whether it can be derived in the sense that it necessarily must look like that, but I can give you the above physical motivation why an action like this is meaningful/worthy to be studied.

Answer 2

One way which I was able to understand this formula is the following:

Let the vector bundle with fibre a vector space $V$ be defined on an open cover $U_\alpha$ and transition maps $g_{\alpha\beta}:U_{\alpha\beta}\to G\subset GL(V)$ (I'm avoiding using representation theory explicitly so let's assume matrix Lie groups). This means if $\sigma_\alpha:U_\alpha\to V$ defines a section on $U_\alpha$ then $(p,\sigma_\alpha)\sim (p,g_{\alpha\beta}\sigma_\alpha)\in U_\beta\times V$.

A connection on this bundle is a Lie algebra valued 1-form. We can define it as collection of Lie algebra $1$-forms on the open cover, so maps $$A_\alpha: U_\alpha\to \Omega^1U_\alpha\otimes\underline{\mathfrak{g}}$$

and make them satisfy some compatibility condition so it's a well defined 1-form.

We can look at the corresponding covariant derivative for this, i.e. $$g_{\alpha\beta}\cdot(d+A_\alpha)\sigma_\alpha= (d+A_\beta)(g_{\alpha\beta}\cdot\sigma_\alpha)$$ should hold.

Expanding this we find $$g_{\alpha\beta}\cdot(d\sigma_\alpha+A_\alpha\otimes\sigma_\alpha) = dg_{\alpha\beta}\otimes\sigma_\alpha + g_{\alpha\beta}\otimes d\sigma_\alpha + A_\beta\otimes g_{\alpha\beta}\sigma_{\alpha}$$ $$g_{\alpha\beta}A_\alpha\otimes\sigma_\alpha = dg_{\alpha\beta}\otimes\sigma_\alpha + A_\beta\otimes g_{\alpha\beta}\sigma_{\alpha}$$ $$A_\alpha\otimes \sigma_\alpha = g_{\alpha\beta}^{-1}dg_{\alpha\beta}\otimes\sigma_\alpha + g_{\alpha\beta}^{-1}A_\beta g_{\alpha\beta}\otimes\sigma_{\alpha}$$ $$A_\alpha = g_{\alpha\beta}^{-1}dg_{\alpha\beta} + g_{\alpha\beta}^{-1}A_\beta g_{\alpha\beta}$$ where the $g_{\alpha\beta}$ can transition across the $\otimes$ since it is just a multiplication by a matrix, not involving derivatives like $A_\alpha$. The only care we took was to respect the order, for example $A_\beta g_{\alpha\beta}\neq g_{\alpha\beta}A_\beta$ since multiplication by matrices is a priori not commutative.

Using $g = g_{\beta\alpha}$ it becomes $$A_\alpha = gA_\beta g^{-1} - g^{-1}dg.$$

Now, the group of gauge transformations is exactly the set of maps $g :M\to G$ corresponding to "changes of basis."

Again, the covariant derivative should be independent of which basis we use. If we consider the same computation above, but seeing $A_\alpha, A_\beta$ not as the connection in different trivializing open sets but as the connection in different choices of gauge (i.e., $A_\alpha=g\cdot A_\beta$), then the we deduce what the action of the gauge group on connections should be, namely, $$gA = gAg^{-1} - g^{-1}dg.$$