I would like to know differential forms representing the cohomology classes of $SU(3)$. I know that there exist a unique bi-invariant form in each class, but I'm not highly motivated by simply putting $R\_g^\*\omega = \omega = L\_g^\*\omega$ and solving it by hand in coordinates. Is there a way to do it less brutally ? On a side note, what is the topology of $SU(3)$, or where can I find an answer ? I have looked in Strickland's [Bestiary of Topological Objects](http://neil-strickland.staff.shef.ac.uk/courses/bestiary/bestiary.pdf), but didn't find the section on unitary groups very enlightening.



De Rham representatives of the cohomology classes in $H^*(SU(3))$
The Poincare polynomial of $SU(3)$ is


$$ (1+t^3)(1+t^5). $$


The cohomology algebra is generated by two elements: a generator $x\_3$ in dimension $3$ and one generator $x\_5$ in dimension $5$, given by the formulas mentioned by Alexander Chervov


$$ x\_k ={\rm Tr}\; (g^{-1}dg)^k,\;\; k=3,5. $$ 


For an explicit description of the generators of the cohomology ring of an arbitrary compact Lie group Example 8.1.15 of [my lecture notes.](http://www.nd.edu/~lnicolae/Lectures.pdf)


