The Symmetric Subalgebra Must Contain Multi-Site Operators
A natural first guess for the generators of the \(\mathbb{Z}_2\) symmetric subalgebra on a spin chain is just the site-wise symmetric operators — one copy of the single-site symmetric algebra at each site. This post shows that guess is badly wrong, via a simple dimension count.
Set-up
Consider a chain of length \(L\) with \(d\)-dimensional qudits. Fix a \(\mathbb{Z}_2\) symmetry acting as
\[U = \bigotimes_{i=1}^L \mathrm{diag}(a_1, \ldots, a_d), \qquad a_i \in \{\pm 1\}\]The on-site Hilbert space decomposes into \(\pm 1\) eigenspaces:
\[\mathcal{H}^{(i)} = \mathcal{H}_+ \oplus \mathcal{H}_-\]with \(p = \dim(\mathcal{H}_+)\) and \(q = \dim(\mathcal{H}_-)\), so \(d = p + q\).
The full chain Hilbert space then splits into even and odd sectors — configurations with an even (resp. odd) number of sites in the \(\mathcal{H}_-\) state:
\[\mathcal{H} = \bigotimes_{i=1}^L \mathcal{H}^{(i)} = \mathcal{H}_{even} \oplus \mathcal{H}_{odd}\]The symmetric subalgebra is \(\mathcal{A}_{\mathbb{Z}_2} \cong \mathrm{End}(\mathcal{H}_{even}) \oplus \mathrm{End}(\mathcal{H}_{odd})\).
The Dimension Count
Site-wise symmetric algebra. At a single site, the \(\mathbb{Z}_2\)-symmetric operators are block-diagonal:
\[\mathcal{A}_{\mathbb{Z}_2}^{(i)} \cong \mathrm{End}(\mathcal{H}_+) \oplus \mathrm{End}(\mathcal{H}_-) \implies \dim = p^2 + q^2\]Tensoring independently across \(L\) sites:
\[\dim(\mathcal{A}_{\mathbb{Z}_2}^{site\text{-}wise}) = (p^2 + q^2)^L\]Full symmetric algebra. Let \(d_e^L = \dim(\mathcal{H}_{even}^L)\) and \(d_o^L = \dim(\mathcal{H}_{odd}^L)\). Two facts:
\[d_e^L + d_o^L = d^L \qquad d_e^L - d_o^L = \mathrm{tr}(U^{\otimes L}) = \mathrm{tr}(U)^L = (p - q)^L\]Solving and squaring:
\[\dim(\mathcal{A}_{\mathbb{Z}_2}) = (d_e^L)^2 + (d_o^L)^2 = \frac{1}{2}\left(d^{2L} + (p-q)^{2L}\right)\]The Ratio Blows Up
We compare the two dimensions:
\[\frac{\dim(\mathcal{A}_{\mathbb{Z}_2})}{\dim(\mathcal{A}_{\mathbb{Z}_2}^{site\text{-}wise})} = \frac{\frac{1}{2}(d^{2L} + (p-q)^{2L})}{(p^2+q^2)^L} \geq \frac{d^{2L}}{2(p^2+q^2)^L} = \frac{1}{2}\left(\frac{d^2}{p^2+q^2}\right)^L\]Now note that
\[\frac{d^2}{p^2 + q^2} = \frac{(p+q)^2}{p^2+q^2} = 1 + \frac{2pq}{p^2+q^2}\]Since \(p, q \geq 1\) for any non-trivial \(\mathbb{Z}_2\) symmetry, we have \(\frac{2pq}{p^2+q^2} > 0\), so the base exceeds 1. Therefore
\[\lim_{L \to \infty} \frac{\dim(\mathcal{A}_{\mathbb{Z}_2})}{\dim(\mathcal{A}_{\mathbb{Z}_2}^{site\text{-}wise})} = +\infty\]Conclusion
The site-wise symmetric algebra grows as \((p^2 + q^2)^L\), but the true symmetric subalgebra grows as \(\sim \frac{1}{2} d^{2L}\). These are exponentially different for large \(L\). The gap
\[\dim(\mathcal{A}_{\mathbb{Z}_2}) - \dim(\mathcal{A}_{\mathbb{Z}_2}^{site\text{-}wise}) \to +\infty\]means there are operators in \(\mathcal{A}_{\mathbb{Z}_2}\) that genuinely can’t be built from single-site symmetric operators alone. Multi-site operators are not optional — they are structurally required.
The natural follow-up question is: what are they? That’s the subject of the next post, where we use Burnside’s theorem to find an explicit generating set.