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As stated in the title, why does a Heisenberg magnet break the $O(3)$ symmetry while degrees of freedom of the underlying spins are $SU(2)$?

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    $\begingroup$ While someone familiar with the Heisenberg model will almost surely know this, it is generally better to make the question self-contained. What is "the $\mathrm{O}(3)$ symmetry, and what is the supposed breaking mechanism? There's no need to have questions be one-liners, and a bit of context makes a question also accessible to people who might not off-hand remember how exactly the Heisenberg model works. $\endgroup$ – ACuriousMind Feb 28 '16 at 11:13
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The Heisenberg model: $$H = \sum_{i \ne j} \mathbf{S}_i\cdot \mathbf{S}_j$$ has an $O(3)$ symmetry group rather than $SU(2)$ even though it is expressed in terms of spin $\frac{1}{2}$ operators $$\mathbf{S}_i = [\frac{\sigma^x_i}{2}, \frac{\sigma^y_i}{2}, \frac{\sigma^z_i}{2}]$$ It is because the Pauli-matrices transform according to the adjoint representation of $SU(2)$, and the adjoint representation of $SU(2)$ is not faithful (rather it is a faithful representation of an $SO(3)$ subgroup) because the element $$g_1 = \begin{pmatrix} -1&0\\ 0& -1 \end{pmatrix}$$ is represented by $1$ on the Pauli matrices, since they transform according $$g_1 \circ \sigma^a_i = g_1^{-1}\sigma^a_i g_1 = \sigma^a_i $$ Thus the largest $SU(2)$ symmetry subgroup of the Heisenberg model is $SO(3)$. However, the Heisenberg model is also symmetric under the parity operator: $$P\circ \sigma^a_i = -\sigma^a_i $$ This operator lies outside $SU(2)$ (it has a determinant of $-1$). It combines with $SO(3)$ to form an $O(3)$ group which is the full symmetry group of the Heisenberg model.

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  • $\begingroup$ Then does the spin nematic break the $SU(2)$ or $O(3)$ symmetry? $\endgroup$ – R.Wigner Feb 29 '16 at 22:29
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    $\begingroup$ @hongchaniyi Spin nematic order breaks O(3) symmetry too. You just need to remember, a spin system does not have SU(2) symmetry in any case. You can talk about SU(2) only for electron systems where the symmetry is fractionalized in some sense. $\endgroup$ – Everett You Feb 29 '16 at 23:22
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    $\begingroup$ @hongchaniyi Indeed, the point symmetry group of a nematic state is $D_{\infty h}$ , a subgroup of $O(3)$. $\endgroup$ – David Bar Moshe Mar 2 '16 at 9:57
  • $\begingroup$ Wouldn't it be more standard to call your operator the time-reversal rather than parity operator, since it flips vectors in spin rather than in real space? (I know this can be a subtle issue, since high-energy and condensed-matter physicists sometimes use different definitions for the time-reversal operator.) $\endgroup$ – tparker Jun 4 '18 at 3:45
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    $\begingroup$ @Quantum spaghettification, yes the group acts on the generators of its Lie algebra (in any representation) by means of the adjoint representation, because the Lie algebra as a vector space forms a basis of the adjoint representation. The infinitesimal version of this action is just the commutation relations. $\endgroup$ – David Bar Moshe Jun 4 '18 at 14:43

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