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Consider the two components spinors (usually Weyl spinors): $\psi = (c_1, c_2)^{\top}$, where $c_i$ are complex numbers. The metric in spinor space is usually defined with the second Pauli matrix $\sigma_y$ (many authors are adding an $i$ factor): \begin{equation}\tag{1} \epsilon = i \, \sigma_y = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}. \end{equation} The spinor scalar product is (in matrix form) \begin{equation}\tag{2} \langle \, \phi, \, \psi \, \rangle = \phi^{\top} \epsilon \, \psi. \end{equation} In particular: \begin{equation}\tag{3} \langle \, \psi, \, \psi \, \rangle \equiv 0. \end{equation} It can be shown that this scalar product is invariant under rotations and Lorentz transformations.

Since $\epsilon^{\top} = -\, \epsilon$, the spinor metric is antisymmetric.

So my question is simple: can we say that the spinor space is symplectic, just from its antisymmetric metric? (like what we say for the classical phase space which is also symplectic, because its metric is also antisymmetric)

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1 Answer 1

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  1. Yes, the vector space $V\cong \mathbb{C}^2$ for the left Weyl representation $\rho:SL(2,\mathbb{C})\to {\rm End}(V)$ is a symplectic vector space with symplectic structure given by the Levi-Civita tensor $\epsilon_{\alpha\beta}$. It is invariant under the group action $$ \langle \rho(g)\phi , \rho(g)\psi \rangle ~=~ \phi^{\top} g^{\top} \epsilon g\psi~=~\det(g)\phi^{\top} \epsilon \psi~=~\langle \phi , \psi \rangle, \qquad g~\in~SL(2,\mathbb{C}), $$ of the double cover $SL(2,\mathbb{C})$ of the restricted Lorentz group.

  2. Recalling that physical left Weyl spinors $\psi^{\alpha}$ are Grassmann-odd, the antisymmetric tensor $\epsilon_{\alpha\beta}$ can be viewed as a metric on a Grassmann-odd super-vector-space.

  3. There is a similar story for the right Weyl/complex conjugate representation $\bar{V}$.

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