9
$\begingroup$

Can you explain me some of the mathematical details of such concept as symmetries? In physics, we have some manifold, and fields are functions on this manifold.

On the one hand, we have symmetries of fields: for example, in the simplest case of spontaneous symmetry breaking group $\mathbb{Z}_{2}$ acts on the algebra of functions on the manifold. But coordinates remain untouched.

On the other hand, there are spatial symmetries in physics, for example Poincare symmetry. In this case, as I understand, Poincare group acts on our manifold (in the mathematical sense: Group action) and somehow induces an action on the algebra of fields (How? Can you explain this construction strictly?).

Also we can consider a situation with both kinds of symmetry:

$$ S=\int \left[\left(\partial_{a} \phi_{1} \right)^{2}+ \left(\partial_{a} \phi_{2} \right)^{2}+ \frac{m^2 \phi_{1}^{2}}{2} + \frac{m^2 \phi_{2}^{2}}{2}\right] $$

In this case we have internal symmetry group $O(2)$: $S[\phi(x)]=S[M\phi(x)]$, $M^{T}M=1$.
It was said to me that in this case physicists write $P\otimes O(2)$, where $P$ is Poincare group, $O(2)$ is acting only on fields, but Poincare group on both space manifold and fields.

So my questions are:

What is the relation between these symmetries and what is the exact construction here? How group acting on space manifold induces action on fields?

Can you explain this strictly in mathematical sense?

$\endgroup$
5
$\begingroup$

Consider a theory of fields $\phi:M\to T$ where $M$ is a manifold, and $T$ is a set. In physics, $T$ is often either a vector space or a manifold. We call $M$ the domain of the theory, and we call $T$ the target space. of the theory. We call a function from $M$ to $T$ a field configuration, and the set of all field configurations is denoted $\mathcal F$.

Let's consider two situations:

Case 1. Let groups $G_M$ and $G_T$ be given. Let $\rho_M$ be an action of $G_M$ on $M$, and let $\rho_T$ be an action of $G_T$ on $T$, then there is a "natural" action of $\rho_\mathcal F$ of $G_M\times G_T$ on $\mathcal F$ given by \begin{align} \rho_\mathcal F(g_M, g_T)(\phi)(x) = \rho_T(g_T)\Big(\phi\big(\rho_M(g_M)^{-1}(x)\big)\Big) \end{align} I'll leave it to you to prove that this is a group action.

Case 2. Let a groups $G$ be given. Let $\rho_M$ be an action of $G$ on $M$, and let $\rho_T$ be an action of $G$ on $T$, then there is a "natural" action of $\rho_\mathcal F$ of $G$ on $\mathcal F$ given by \begin{align} \rho_\mathcal F(g)(\phi)(x) = \rho_T(g)\Big(\phi\big(\rho_M(g)^{-1}(x)\big)\Big) \end{align} I'll leave it to you to prove that this is in fact a group action.

Now, let us phrase your question in the following way:

In either of the above cases, is their a sense in which $\rho_M$ induces $\rho_T$?

The answer, as far as I am aware, is that it depends on the context and on what you mean by "induced." Let's consider the example you give in your statement of the question.

Example. An $\mathrm{O}(2)$ vector field on $\mathbb R^{3,1}$.

We have \begin{align} M = \mathbb R^{3,1}, \qquad T = \mathbb R^2, \qquad G_M = \mathrm{P}(3,1), \qquad G_T = \mathrm{O}(2) \end{align} where $\mathrm{P}(3,1)$ is the Poincare group in four dimensions. In this case, one often takes $\rho_M$ and $\rho_T$ to be \begin{align} \rho_M(\Lambda,a)(x) = \Lambda x+a, \qquad \rho_T(R)(v) = Rv \end{align} Notice that this falls under Case 1 above. In this case, there is no "canonical" relationship (as far as I am aware) between the Poincare group and $\mathrm{O}(2)$, so there is no canonical sense in which $\rho_M$ induces $\rho_T$.

However, consider the following example:

Example. An $\mathrm{SO}(3)$ $2$-tensor field on $\mathbb R^{3}$.

We have \begin{align} M = \mathbb R^{3}, \qquad T = T_2(\mathbb R^3), \qquad G = \mathrm{SO}(3) \end{align} where $T_2(\mathbb R^3)$ is the vector space of $2$-tensors on $\mathbb R^3$. Then there is a natural action of $G$ on $M$ given by \begin{align} \rho_M(R)x = Rx \end{align} Since $2$-tensors on $\mathbb R^3$ can be described as bilinear maps functions $S:\mathbb R^3\times \mathbb R^3 \to \mathbb R$, there is also a natural action of $G$ on $T$ given by \begin{align} \rho_T(S)(x_1, x_2) = S(R^{-1}x_1, R^{-1} x_2) \end{align} which can also be written as \begin{align} \rho_T(S)(x_1, x_2) = S(\rho_M(R)^{-1}x_1, \rho_M(R)^{-1} x_2) \end{align} so, in this case, there is a sense in which $\rho_M$ has induced $\rho_T$.

| cite | improve this answer | |
$\endgroup$

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.