Einstein notation with three indexes

443 Views Asked by At

I met in a book a mention of Sklyanin algebra for $(S_0,S_1,S_2,S_3)$: $$\{S_0,S_{\alpha}\}=\varepsilon_{\alpha\beta\gamma}S_{\beta}S_{\gamma}(J_{\beta}-J_{\gamma})$$ $$\{S_{\alpha},S_{\beta}\}=\varepsilon_{\alpha\beta\gamma}S_{0}S_{\gamma}$$ where $\alpha,\beta,\gamma\in\{1,2,3\}$ and $J_1,J_2,J_3$ are some constants. My question is, how can there be three indexes in one expression? Can I consider them like Einstein summing but with three components?

1

There are 1 best solutions below

1
On BEST ANSWER

In the first tensor equation, it appears $\alpha$ is the free index whereas $\beta,\gamma$ are dummy indices for Einstein summation notation (where repeated indices implies summation over said indices). Usually we use raised and lowered indices in Einstein notation, but whatever. This means for each of $\alpha=1,2,3$ we sum over all possible values of $\beta$ and $\gamma$ (from $\{1,2,3\}$?) on the right-hand side. For example if $\alpha=1$ we get

$$ \begin{array}{ccccccc} \{S_0,S_1\} = & \phantom{+}\,\varepsilon_{111}S_1S_1(J_1-J_1) + \varepsilon_{112}S_1S_2(J_1-J_2) +\varepsilon_{113}S_1S_3(J_1-J_3) \\ & +\, \varepsilon_{121}S_2S_1(J_2-J_1) + \varepsilon_{122}S_2S_2(J_2-J_2) +\varepsilon_{123}S_2S_3(J_2-J_3) \\ & +\,\varepsilon_{131}S_3S_1(J_3-J_1) + \varepsilon_{132}S_3S_2(J_3-J_2) +\varepsilon_{133}S_3S_3(J_3-J_3) \end{array} $$

So there are nine summands on the right-hand side of each equation (for all values of $\beta$ and $\gamma$) and there are three scalar equations (for all values of $\alpha$). Of course, $J_\beta-J_\gamma=0$ if $\beta=\gamma$ so the summands along the "diagonal" (as I have represented them in an array) vanish.

In the second tensor equation, it appears $\alpha$ and $\beta$ are free variables while $\gamma$ is the dummy variable on the right. Thus for example if $\alpha=1$ and $\beta=2$ then the equation reads

$$ \{S_1,S_2\}=\varepsilon_{121}S_0S_1+\varepsilon_{122}S_0S_2+\varepsilon_{123}S_0S_3. $$

So the second tensor equation splits into nine possible scalar equations, each involving a sum of three terms on its right-hand side. This is my interpretation, anyway.