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Malkoun
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Lie algebras, root systems and qubits

This post is about some concepts I am experimenting with. They are related to the Atiyah problem on configurations. They kind of mix Lie algebras and qubits. Given a compact (say semisimple) Lie group $G$ with Lie algebra $\mathfrak{g}$, we choose a compact torus $T$ of $G$, with Lie algebra $\mathfrak{t}$. Each root of $\mathfrak{g}$ with respect to the complexification of $\mathfrak{t}$ restricts to an element in $\mathfrak{t}^*$.

Consider $\mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta$, where $$ \Delta = \{ \operatorname{ker}(\alpha \otimes \mathbf{1}) ; \alpha \in \Phi \} $$ where $\Phi$ is the set of roots of $\mathfrak{g}$ (with each root restricted to $\mathfrak{t}$) and $\mathbf{1}$ denotes the identity on $\mathbb{R}^3$. Thus each $\alpha \otimes \mathbf{1}$ is a linear map from $\mathfrak{t} \otimes \mathbb{R}^3$ to $\mathbb{R}^3$.

Let us introduce the notation $$ \mathcal{C} = \mathfrak{t} \otimes \mathbb{R}^3 \setminus \Delta. $$

Given $\mathbf{x} \in \mathcal{C}$ and $\alpha \in \Phi$, note that $(\alpha \otimes \mathbf{1})(\mathbf{x}) \in \mathbf{R}^3 \setminus \{\mathbf{0}\}$. But $\mathbf{R}^3$ is isomorphic to the Lie algebra $\mathfrak{su}(2)$ (or actually, to be closer to physicists, we consider $i\mathfrak{su}(2)$ instead, which consists of trace-free hermitian $2$ by $2$ matrices), via the map: $$ x^a \mapsto x^a \sigma_a, $$ for $a = 1, \ldots, 3$, where $\sigma_1$, $\sigma_2$ and $\sigma_3$ are the Pauli matrices. Thus $(\alpha \otimes \mathbf{1})(\mathbf{x}).\mathbf{\sigma}$ is a trace-free hermitian $2$ by $2$ matrix, where $\mathbf{\sigma}$ is the $3$-vector of Pauli matrices. Let $\psi_{\alpha}(\mathbf{x})$ denote a normalized eigenvector of it in $S^3 \subset \mathbb{C}^2$ with respect to its positive eigenvalue. Note that $\psi_{\alpha}(\mathbf{x})$ is only well defined up to a phase factor. Nevertheless, the map from $\mathcal{C}$ to $\mathbb{C}P^1$ which maps $\mathbf{x}$ to $[\psi_{\alpha}(\mathbf{x})]$ is a well defined smooth map, which we will denote by $\hat{\psi}_\alpha$.

We can think of $\psi_\alpha(\mathbf{x})$ as a qubit. This allows us to do the following. If we choose a direction, say $\nu \in S^2$, which for simplicity, I will assume it is $(0, 0, 1)^T$ (the "up" direction), then at each $\mathbf{x} \in \mathcal{C}$, a root $\alpha$ will be positive with probability amplitude $u$ (and thus probability $|u|^2$) and negative with probability amplitude $v$ (and thus probability $|v|^2$), with respect to the chosen direction $\nu$, where $\psi_\alpha(\mathbf{x}) = (u, v)^T \in S^3 \subset \mathbb{C}^2$.

I will not provide a complete description here, but if $G = SU(n)$ and if we form the $|\Phi|$-qubit $$ \psi(\mathbf{x}) = \bigotimes_{\alpha \in \Phi} \psi_{\alpha}(\mathbf{x}), $$ then the Atiyah-Sutcliffe determinant (related to the Atiyah problem on configurations) can then be thought of as the probability amplitude associated to an irreducible $1$-dimensional summand of the above $|\Phi|$-qubit space, with the latter thought of as a representation space of $SU(2)$. Thus the absolute value squared of the Atiyah-Sutcliffe determinant (properly normalized) is the probability that the quantum state $\psi$ is in that $1$-dimensional subspace after "measurement". The Atiyah-Sutcliffe conjecture 2 gives thus a positive lower bound on this probability. The Atiyah-Sutcliffe conjecture 1 (non-vanishing of the Atiyah-Sutcliffe determinant) can then be interpreted as the following statement: there is no $\mathbf{x} \in \mathcal{C}$ for which the corresponding quantum state $\psi(\mathbf{x})$ is forbidden to lie in this $1$-dimensional subspace. Stated this way, this seems quite plausible (in my opinion)!

I would like to pursue this line of thought, bringing in qubits and QM to Lie algebras, with the aim of applying such ideas to the Atiyah problem on configurations. I guess I wanted to share my idea, and wonder if similar ideas were considered before. I know there exist quantum groups, but from what I understand, it is the Lie bracket which gets deformed, so it is not the same as what I am considering here.

Thoughts, remarks and comments are welcome. References to related ideas in the literature are welcome too.

Malkoun
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