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This question is related to requirement that the gauge group of a gauge theory be a direct product of compact simple groups and $U(1)$ factors but is not the same as, for example, this question (though related as described below).

When looking to build the kinetic term of a gauge theory, the demand that the Lagrangian be real and Lorentz invariant implies the lowest order term we can write down using only the field strength $F^{\alpha}_{\mu\nu}$ (using $\alpha,\beta,\ldots$ for gauge group indices) is of the form $$ g_{\alpha\beta}F^{\alpha}_{\mu\nu}F^{\beta\mu\nu} $$ where $g_{\alpha\beta}$ is a real matrix which we may take to be symmetric.

As described in this answer and Weinberg Vol2, in order to conclude that the gauge group must be a product of compact simple and $U(1)$ factors we must ague both that $g_{\alpha\beta}$ satisfy the invariance condition $g_{\alpha\beta}C^\beta_{\gamma\delta}=-g_{\gamma\beta}C^\beta_{\alpha\delta}$ (so $g_{\alpha\beta}$ is proportional to the Killing form of the gauge group) and also that $g_{\alpha\beta}$ must be positive-definite.

The former of these follows from gauge invariance and does not concern me here. My question: The claim is that the positive definiteness of $g_{\alpha\beta}$ follows from unitarity and the canonical quantization procedure. Can anyone make explicit how this follows?

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The shortest route is to Wick rotate. In the Euclidean setting, the integration measure is $$ e^{-S[A]}\mathrm dA,\qquad\text{with}\qquad S[A]=\frac{1}{g^2}\int \langle F,F\rangle $$

If the scalar product is not positive definite, then the action does not decay for large field configurations, and the path-integral does not converge. So the QFT does not even exist.

In the Lorentzian setting the philosophy is really the same. Recall that in the path-integral we send the time direction in a slightly imaginary direction (see this PSE post), and so you still need the imaginary part of the action to have the appropriate decay properties. This can all be traced back to the assumption that the Hamiltonian is hermitian and bounded from below, so if a QFT with non-positive-definite metric exists, it cannot have a healthy Hilbert space with well-defined Hamiltonian.

There is an interesting loophole though. If the classical phase space is finite-dimensional, then the path-integral converges even if $S$ does not decay. This plays an important role in topological gravity, where we the gauge group is Poincare (which is not simple so its Killing form is not positive definite). The theory is well-defined even though $\langle\cdot,\cdot\rangle$ is not a norm. See the seminal work of Witten for more details regarding this point.

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A different (and perhaps more convincing) argument is the following. Instead of looking at Lagrangians, you can construct the theory of spin-1 particles from the bottom-up. We look at the most general $S$-matrix of such particles, and impose that it leads to a self-consistent theory. A very nice account of this construction is Scharf, Quantum gauge theories, a true ghost story. In section 3.2 the author proves that the coupling of these particles necessarily involves a set of constants $f_{ijk}$ which are completely anti-symmetric and satisfy the Jacobi identity. (This also follows from Weinberg's soft theorems, see e.g. Schwartz's Quantum Field Theory and the Standard Model, section 27.5.2).

It is a classic result of the theory of Lie algebras that the set $f_{ijk}$ satisfying Jacobi and being anti-symmetric in its three indices defines a reductive Lie algebra, and vice-versa. So this proves that self-consistent theories of interacting spin-1 particles are classified by positive-definite Killing forms.

Regarding the loophole above, topological theories have no particles, so this is how they evade the constraints from the $S$-matrix.

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  • $\begingroup$ I accept your arguments as convincing, but can you comment briefly on the relation to unitarity? This is specifically what Weinberg cites. $\endgroup$ Jan 12, 2021 at 19:47
  • $\begingroup$ @RichardMyers Consistent interactions of spin-1 particles must be labelled by anti-symmetric structure constants. This follows from consistency of the $S$-matrix. In a non-unitary theory you don't have an $S$-matrix (or this matrix is not constrained by unitarity), so the argument breaks down. $\endgroup$ Jan 16, 2021 at 19:17

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