# $\left(H^\dagger H\right)^2$ is invariant under $U(1)\times SU(2)$?

Is it true that $\left(H^\dagger H\right)^2$ is invariant under $U\left(1\right) \times SU\left(2\right)$ where $H$ is the Higgs field $(1,2,1/2)$?

Does this invariance imply that its hypercharge is invariant under $U\left(1\right)$ and its spin is invariant under $SU\left(2\right)$? .

$$H = [H_+, H_0]$$

$$H_+ = [H_-]$$

but

$$H_0 = [?]$$

$$H^\dagger H = [H_-][H_+] + [?][H_0]$$

Yes, $(H^\dagger H)^2$ is invariant under $SU(2)\times U(1)$ because even without the second power, $H^\dagger H$ is invariant under it. By that, we mean $$\sum_{i=1}^2 H_i^* H_i$$ Of course that it's invariant under $U(1)$ because the $U(1)$ charges of $H^\dagger$ and $H$ are opposite in sign and add up to zero. Note that the transformation of a charge-$Q$ field is $F\to F\exp(iQ\lambda)$.
The invariance under $SU(2)$ is also self-evident because $\sum_i z^*_i z_i$ is exactly the bilinear (with one asterisk) invariant that defines the unitary groups.
No, the invariance of $H^\dagger H$ or its square does not mean that the hypercharge of $H$ itself is zero or isospin is zero. Moreover, the OP seems to confuse the spin and the isospin.