# How can I tell that a Lagrangian has an $SU(2)\times SU(2)$ symmetry?

this is a very basic question and it probably has a very simple answer.

I was reading through some handouts when I came over something that I did not understand. One considered the simple Lagrangian

$$\mathscr{L}=\partial_\mu\phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi - \lambda\left(\phi^\dagger\phi\right)^2$$ where $\phi$ is a complex doublet. The author then said that the symmetry of this Lagrangian is $SU(2)\times SU(2)$. However, I thought it was just $SU(2)$ or $U(1)\times U(1)$? I then googled this and found that the Lorentz group is isomorphic to $SU(2) \times SU(2)$, which I guess is one explanation. However, I was wondering if one could show that the Lagrangian is invariant under $SU(2) \times SU(2)$ by acting with some transformation, without taking the shortcut with Lorentz invariance?

Any help would be greatly appreciated (:

• For at least one $SU(2)$ group notice that $\phi^\dagger \phi \to \phi^\dagger U^\dagger U \phi$ when you take $\phi \to U \phi$. So if $U$ is unitary the lagrangian will be invariant. Jun 30, 2016 at 11:28
• Excellent. But what does it mean that the symmetry is $SU(2)*SU(2)$? Jun 30, 2016 at 11:39
• Also the Lorentz group is not actually isomorphic to $SU(2)\times SU(2)$ as Lie groups because the former is non-compact whereas the latter is compact. At the Lie algebra level there is a trick using complexification where you can split the Lorentz groups algebra into two commuting $su(2)$'s. Jun 30, 2016 at 11:40
• $\uparrow$ Complex doublet in what sense? E.g. as a $SU(2)$ doublet representation? Jun 30, 2016 at 11:44
• $\phi = (\phi_1,\phi_2)$ with $\phi_1$ and $\phi_2$ being complex fields. Jun 30, 2016 at 11:46

You already got your answer, all right, several times over, but I will emphasize the central puzzle of your question which you only got indirect answers for, connected to the peculiar special structure of SO(4). Any self-respecting text introducing the standard model more or less has it. I'll skip all superfluous issues like lagrangian terms, the U(1)s, etc... and stick to the invariance of the bilinear of scalars, at the heart of your puzzlement.

I suspect you are asking how is the bilinear $$\phi^\dagger \cdot \phi$$ invariant under two different, commuting SU(2)s instead of the familiar gauged SU(2) left-acting on the complex vector, $$$$\phi \equiv\begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix} .$$$$ First you need to recall that SU(2) is pseudoreal, i.e. the conjugate representation $$\tilde{\phi}\equiv i\tau_2 \phi^*$$ is equivalent to this fundamental one, that is, acting on $$$$\tilde{\phi} = \begin{pmatrix} \phi^*_{2}\\- \phi^*_1 \end{pmatrix}$$$$ produces the same transformation on all four components of $$\phi$$ , real and imaginary pieces.

Now consider the 2x2 complex matrix with columns $$\tilde{\phi}$$ and $$\phi$$, respectively, so $$\Phi\equiv \frac{1}{\sqrt{2}} [ \tilde{\phi} ~,~ \phi ]$$, so $$\Phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi^*_{2}& \phi_{1} \\ -\phi^*_{1}& \phi_{2} \end{pmatrix} ~,$$ and check that $$\Phi^\dagger \Phi = (\phi^\dagger \!\cdot \phi)~ 1\!\!\!1 ~/~2 ~,$$ so $$\operatorname {Tr} \Phi^\dagger \Phi = \phi^\dagger\! \cdot \phi$$.

Now, left multiplication of $$\Phi$$ by a unitary SU(2) rotation leaves the matrix bilinear of $$\Phi$$s invariant, and, a fortiori, its trace invariant.

Significantly, however, right multiplication by another, independent SU(2) which knows nothing about the left one, scrambles the two columns of $$\Phi$$ with each other, preserving, however their left-SU(2) transformation properties, since, as we saw, each column of $$\Phi$$ amounts to the same outcome when left-SU(2) transformed. It is evident, but you may convince yourself by a right transformation, a left transformation, and then the inverse right transformation--you will be left with the original left transformation.

Even though the $$\Phi$$ bilinear is not right-SU(2) invariant, by the cyclicity of the trace, its trace is. So all $$\phi^\dagger \cdot \phi$$ bilinears are both left, but also right SU(2) invariant, and so are all lagrangian kinetic and potential terms you'd construct.The right-invariance, however, will be spoiled by coupling to gauge fields, as you might check.

Answer of this question is quite subtle. First let us consider the most general Higgs potential which is renormalizable and invariant under $SU(2)_{L}\otimes U(1)_{Y}$ gauge transformations, which has the form $$V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2}$$ Where $$\phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+i\phi_{2} \\ \phi_{3}+i\phi_{4} \end{pmatrix}$$ In terms of $\phi_{i}(i=1,2,3,4)$, Higgs potential can be expressed as $$V = \frac{\lambda}{4}(\phi^{2}_{1}+\phi^{2}_{1}+\phi^{2}_{3}+\phi^{2}_{4}-2\mu^{2})^{2}$$ If we define $$\Phi=\begin{pmatrix} \phi_{1}\\ \phi_{2}\\ \phi_{3}\\ \phi_{4}\\ \end{pmatrix}$$ Then, Higgs potential is invariant under rotations of the four fields which lead to $SO(4)$ as the global symmetry group. This group is isomorphic to $SU(2)\otimes SU(2)$, because both have the same Lie algebra. This symmetry is global and it does not necessary to introduce gauge fields.

When the symmetry is broken, the scalar field $\phi_{4}$ get a v.e.v different from zero, and it can be redefined as follows $\phi_{4}=H+v$. where H gets its mass and is called the Higgs particle. Moreover, it has a v.e.v equal to zero. This field is a physical degree of freedom and its mass is proportional to the $λ$ parameter which is unknown in the model. The other scalar fields remain massless. They are the would-be Goldstone bosons and correspond to the degrees that the gauge fields ‘eat‘ in order to get mass or longitudinal component. The Higgs potential can be written as a function of the new fields as follows $$V = \frac{\lambda}{4}(\phi^{2}_{1}+\phi^{2}_{1}+\phi^{2}_{3}+H^{2}+2Hv+v^{2}-2\mu^{2})^{2}$$ In this new potential the global symmetry is broken to $SO(3)$, which only rotates three scalar fields. It is isomorphic to $SU(2)_V$, the diagonal part of $SU(2)\otimes SU(2)$. Which is also known as Custodial symmetry.

• Just as note on an otherwise nice answer, $SU(2)$ and $SO(3)$ have the same algebra but aren't isomorphic as groups. Jun 30, 2016 at 18:03

After a bit of discussion I believe there is actually a $SU(2)\times SU(2)$ symmetry in a sense.

So in principle there is a $U(2)$ symmetry if $\phi=(\phi_1,\phi_2)^T$, $\phi^\dagger=(\phi_1^*,\phi_2^*)$ and the lagrangian $$\mathscr{L}=\partial_\mu \phi^\dagger\partial^\mu \phi-m\phi^\dagger\phi-\lambda(\phi^\dagger\phi)^2,$$ simply sent $\phi\to U\phi$, for any unitary matrix $U$.

But what we can try do is to write the complex scalar fields $\phi_i=\Re(\phi_i)+i\Im(\phi_i)$ in terms of real scalar fields.

Lets say to tidy up notation $\phi_1=\eta_1+i\eta_2$ and $\phi_2=\eta_3+i\eta_4$, where the $\eta_i$ are real scalar fields.

Then $$\phi^\dagger\phi=\eta_1^2+\eta_2^2+\eta_3^2+\eta_4^2$$ and $$\partial_\mu \phi^\dagger\partial^\mu\phi=\partial_\mu\phi_1^*\partial^\mu \phi_1+\partial_\mu\phi_2^*\partial^\mu \phi_2= \sum_i\partial_\mu\eta_i\partial^\mu\eta_i$$

So in fact the lagrangian becomes

$$\mathscr{L}=\sum_i\partial_\mu\eta_i\partial^\mu\eta_i-m(\eta_1^2+\eta_2^2+\eta_3^2+\eta_4^2)-\lambda(\eta_1^2+\eta_2^2+\eta_3^2+\eta_4^2)^2$$

Or if we collect $\eta=(\eta_1,\eta_2,\eta_3,\eta_4)^T$ then we have:

$$\mathscr{L}=\partial_\mu \eta^T\partial^\mu \eta-m\eta^T\eta-\lambda(\eta^T\eta)^2$$

This lagrangian of real scalar fields is $SO(4)$ or ($O(4)$) invariant, of which I believe the universal cover is $SU(2)\times SU(2)$. I don't know all the details of representation theory, but I am led to believe, after reading in Weinberg's book anyway, that one can replace a group of symmetries by it's universal cover, in a similar way $SO(3)$ is replaced by $SU(2)$ in quantum mechanics.

There's link here on mathstackexchange about $SO(4)$ and it's relation to $SU(2)\times SU(2)$ and also there's a link in the comments, whereby I think as matrix groups, not lie groups $SO(4)\cong SU(2)\otimes SU(2)$.

Physically though I think the extra symmetry comes about because initially we said that $\phi_1=\eta_1+i\eta_2$ and $\phi_2=\eta_3+i\eta_4$, but equally well we could define $\phi_1=\eta_1 +i\eta_4$ and $\phi_2=\eta_2 +i\eta_3$, or any other variant, and the lagrangian will look the exact same.

Again if you check that link on mathstack that kennytm linked, or for convenience here, the relation of the above groups seems to appear in studies of entanglement in quantum computing.

Maybe another answer or comments could fill in some gaps that I have left.

If the field is a simple complex scalar field, than the symmetry is just $U(1)$. For a higher symmetry, $\phi$ need to be higher dimensional too, for instance you can add a vector index $\phi_i$ with $i=1,2$ for simplicity, which means that you add an additional complex field. If these two fields interact, you can have two cases now:

Each field has a $U(1)$ symmetry, and the interaction terms, e.g. $\lambda_{12}|\phi_1|^2 |\phi_2|^2$, respect the $U(1)$ symmetry. However, the fields have different parameters like masses and self-coupling constants. Then you have separate $U(1)$ symmetries and the overall symmetry of your Lagrangian is $U(1)\times U(1)$.

$$\mathscr{L}=\partial_\mu\phi_1^\dagger \partial^\mu \phi_1 - m_1^2 \phi_1^\dagger \phi_1 - \lambda_1\left(\phi_1^\dagger\phi_1\right)^2+\partial_\mu\phi_2^\dagger \partial^\mu \phi_2 - m_2^2 \phi_2^\dagger \phi_2 - \lambda_2\left(\phi_2^\dagger\phi_2\right)^2+\lambda_{12}|\phi_1|^2 |\phi_2|^2$$

In case 2, the other parameters are the same for both fields, so the symmetry is $SU(2)$, because you can rotate them with the same phase.

$$\mathscr{L}=\partial_\mu\phi_1^\dagger \partial^\mu \phi_1 - m^2 \phi_1^\dagger \phi_1 - \lambda\left(\phi_1^\dagger\phi_1\right)^2+\partial_\mu\phi_2^\dagger \partial^\mu \phi_2 - m^2 \phi_2^\dagger \phi_2 - \lambda\left(\phi_2^\dagger\phi_2\right)^2+\lambda|\phi_1|^2 |\phi_2|^2$$

The Lagrangian of the OP is apparently of this type, where the inner products in the space of the two fields is not written explicitly.

• Thank you for your insight (: However, I still do not understand why we say that the Lagrangian has an $SU(2)*SU(2)$ symmetry? Jun 30, 2016 at 11:44
• the lagrangian you wrote apparently has this vector index I am talking about, just not written explicitly. So it is a short hand notation for my case 2, you have two complex scalar fields, and all interaction terms have the same coupling constants. Jun 30, 2016 at 11:54
• @ Noldig, How you gonna show that $U(1)\times U(1)$ is the same as $SU(2)\times SU(2)$?
– Mass
Jun 30, 2016 at 15:08
• where did I ever claim that? Jul 1, 2016 at 9:10