What does it mean to use levi civita symbol with Poisson brackets in this way 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. 
 A: 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
$$
\{L_1,A_3\}
= \epsilon_{13c}A_c
= \epsilon_{131}A_1+\epsilon_{132}A_2+\epsilon_{133}A_3
= 0A_1+(-1)A_2+0A_3
=-A_2.
$$
A: 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.
