Neither of the references linked in the comments seem to solve the OP's question. Gabriel's theorem says that the indecomposables correspond to positive roots. The way this correspondence works is as follows: you take the positive root you are interested in, and express it as a sum of simple roots (with multiplicities). The multiplicity of the simple root corresponding to node $i$ tells you the dimension of the vector space at node $i$.

However, this still doesn't explicitly construct a representation! All it tells you is the dimension of the underlying vector spaces, without telling you about the linear maps between them. A crucial piece of additional information is that in Dynkin type, if there is an indecomposable representation with dimension vector $\vec d$, then a generic representation of that dimension vector will be indecomposable. So, one answer is to say "take a generic representation of that dimension vector".

For more detail, I suggest the book of Gabriel and Roiter. Its section on hereditary algebras (several chapters in the middle) can be read without reference to what comes before or after, and goes into a lot of detail.

Edited to add: oh, another thing: for Dynkin quivers, there is a way to construct any indecomposable by applying a sequence of "reflection functors" starting from a simple indecomposable, but the sequence of reflections one would have to apply to get to the highest root of $E_8$ is long enough that it probably wouldn't be especially useful to work this out.

only thosehave been classified. Have you tried Google to begin with? What about Quiver (mathematics) $\endgroup$ – Alex Degtyarev Jun 3 '14 at 18:08