In Weinberg's QFT Volume 3 book on Supersymmetry, he presents his own proof of the Coleman-Mandula theorem. As part of the proof, he proves that the only possible internal symmetry generators must form a direct sum of compact semi-simple Lie algebras and U(1) algebras. Label these internal symmetry generators $B_{\alpha}$. Their action on multi-particle states are as follows, using Weinberg's notation: $$B_{\alpha}|pm,qn,...\rangle=\sum_{m'}(b_{\alpha}(p))_{mm'}|pm',qn,...\rangle+\sum_{n'}(b_{\alpha}(q))_{nn'}|pm,qn',...\rangle+...$$

Where $b_{\alpha}(p)$ are finite Hermitian matrices which define the action on single particle states. The result can be easily proven for the single particle matrices, and the remainder of the proof is to show that there is an isomorphism between the $b_{\alpha}(p)$ and the $B_{\alpha}$.

Both of these have the same commutation relations: $$[B_{\alpha},B_{\beta}]=iC^{\gamma}_{\alpha\beta}B_{\gamma}$$ $$[b_{\alpha}(p),b_{\beta}(p)]=iC^{\gamma}_{\alpha\beta}b_{\gamma}(p)$$

This provides a homomorphism between the two, i.e. $f:b_{\alpha}(p)\to B_{\alpha}$.

Weinberg states: "For it $(f)$ to be an isomorphism would require that whenever $\sum_{\alpha}c^{\alpha}b_{\alpha}(p)=0$ for some coefficients $c^{\alpha}$ and momentum $p$, then $\sum_{\alpha}c^{\alpha}b_{\alpha}(k)=0$ for all momenta $k$, which is equivalent to the condition $\sum_{\alpha}c^{\alpha}B_{\alpha}=0$."


Why is this a sufficient condition for there to be an isomorphism between the two generators?

For reference see section 24.B of Weinberg's Quantum theory of fields.

  • $\begingroup$ It would help if you quoted in your question what the "it" in the Weinberg quote refers to, i.e. what the actual isomorphism is supposed to be. $\endgroup$
    – ACuriousMind
    Feb 12 '19 at 18:54
  • $\begingroup$ Weinberg himself does not specify the form of the isomorphism. I added what I think he is referring to. $\endgroup$
    – fewfew4
    Feb 12 '19 at 19:14

Let us instead look at the inverse map $g_p: B_\alpha \mapsto b_\alpha(p)$. The claim we want to prove is that $g_p$ is an isomorphism for all $p$.

  1. A map between Lie algebras ($g_p$) is an isomorphism if and only if it is an isomorphism of vector spaces and it respects the Lie bracket. You already know that it respects the Lie bracket because the $b$ and $B$ have the same commutation relations and you know it is linear, i.e. a homomorphism, because we defined it only on the generators $B$ and are implicitly extending it linearly to the rest of the algebra.

  2. A homomorphism of vector spaces is an isomorphism if and only if it is both surjective and injective. You already know it is surjective because the image of $g_p$ is generated by the images of the $B_\alpha$, which are all of the $b_\alpha(p)$, and since the $b_\alpha(p)$ are the generators of their algebra, the image is the whole algebra.

  3. A homomorphism of vector spaces is injective if and only if its kernel is trivial, i.e. contains only the zero vector. An element $c^\alpha B_\alpha$ is in the kernel of $f$ if and only if $g_p(c^\alpha B_\alpha) = c^\alpha g_p(B_\alpha) = c^\alpha b_\alpha(p) = 0$. So $g_p$ is only injective for all momenta $p$ if $c^\alpha b_\alpha(p) = 0$ for that single $p$ implies $c^\alpha B_\alpha = 0$.

  4. By definition, we have that $B_\alpha$ acts as a sum of actions of $b^\alpha(k_i)$ for some collection of momenta $k_i$. So if $c^\alpha b_\alpha(p) = 0$ for a fixed $p$ implies $c^\alpha b_\alpha(q) = 0$ for all $q$, then it implies in particular that all $c^\alpha b_\alpha(k_i) = 0$, and therefore $c^\alpha B_\alpha = 0$, which in turn shows $g_p$ is injective, which in turn means $g_p$ is an isomorphism of Lie algebras, which is what we wanted to show.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.