What does it mean to use levi civita symbol with Poisson brackets in this way

Question

I'm doing some studies in mathematical methods for physics and I came across something that I don't really understand. I have only been using the $\epsilon_{ijk}$ when I cross some vectors or operators etc... But what does it mean when you write something like this, with Poisson brackets we have that

$$ \{L_a,A_b \}=\epsilon_{abc}A_c $$

Maybe someone can explain what this means because the material I got for the course just assumes that I know this.

Answer 1

From the context, it seems that the indices $a$, $b$, $c$ can only take the values 1, 2 and 3, and then the formula means that $\{L_1,A_1\}=0$, $\{L_1,A_2\}=+A_3$, $\{L_1,A_3\}=-A_2$, and so on.

(The positive sign if $abc$ come in the right cyclic order, negative sign if the wrong order, or zero if there is repeated index.)

In detail: summation over $c$ is understood, so for example $$ \{L_1,A_3\} = \epsilon_{13c}A_c = \epsilon_{131}A_1+\epsilon_{132}A_2+\epsilon_{133}A_3 = 0A_1+(-1)A_2+0A_3 =-A_2. $$

Answer 2

By convention, the repeated index, $c$, on the right hand side is summed over the 3 spatial dimensions. The equation is a short way of writing 9 equations, which can be arranged in 3 groups of 3 equations.

Consider when $a=b=1$. Then $\epsilon_{11c}=0$ for all 3 values of $c$. Thus the equation contains the fact that $\{L_1,A_1\}=0$. Similarly for $a=b=2$ and for $a=b=3$.

Consider when $a=1$ and $b=2$. Then $\epsilon_{12c}$ is nonzero only when $c=3$. Since $\epsilon_{123}=1$, the equation contains the fact that $\{L_1,A_2\}=A_3$. Similarly for 2 other cases.

Consider when $a=2$ and $b=1$. Then $\epsilon_{21c}$ is nonzero only when $c=3$. Since $\epsilon_{213}=-1$, the equation contains the fact that $\{L_2,A_1\}=-A_3$. Similarly for 2 other cases.

We say that when the subscripts are an even permutation of 1,2,3 the Levi-Civita symbol is just one. When the subscripts are an odd permuation (like 213), the symbol is -1. Otherwise, it is zero.