I would like to know differential forms representing the cohomology classes of $SU(3)$. I know that there exist a unique biinvariant 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, but didn't find the section on unitary groups very enlightening.

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. 


Relying on the nice answer by Liviu Nicolaescu on the particular question, let me comment on a little more general setup. My point is that "many" things (cohomology classes, invariant forms/vector fields, invariant polynoms, etc...) can be written very explicitly and downtoearth for classical groups. Such technique is standard in math.physics, but to the best of my knowledge is not presented systematically in basic math. textbooks on Lie groups, where it should be (imho). Claim "Everything" (cohomology classes, invariant forms/vector fields, invariant polynoms, etc...) for A,B,C,D groups can be written as traces, or determinants or other linear algebra expressions applied to some explicitly given matrices ("Lax matrices"). "Examples first" Consider the group GL(2) embed it into Mat(2) so we have natural coordinates on this linear space: "x,y,z,u". Consider their deRham differential, i.e. 1forms: "dx,dy,dz,du". And define the following matrices: $$ M=\begin{bmatrix} x & y \\\\ z & u \end{bmatrix}, ~~~dM= \begin{bmatrix} dx & dy \\\\ dz & du \end{bmatrix} $$ Pay attention on some psychologically important detail: the elements of these matrices are functions and 1forms, but NOT complex numbers. Lemma 1 (Invariant 1forms): take any 2x2 number valued matrix "K", then $Tr(KM^{1}dM)$  left invariant 1form on the group GL(2); $Tr(KdM M^{1})$  right invariant 1form on the group GL(2). Changing matrix "K" we obtain the whole space of left/right invariant 1forms. The map K>1forms correspond to identification of dual to Lie algebra with invariant forms. Proof: the idea is very simple action of group is M> GM, so when you have (GM)^{1}dGM=, (M)^{1}(G)^{1}GdM=(M)^{1}dM. I hope you would agree that these are very explicit coordinatewritten expressions which "everyone can understand". Lemma 2 (Cohomology classes) The group GL(2), is contractible to U(2), not SU(2) and has basic generators of cohomology classes of degrees 1,3. The generators of cohomology classes can be written explicitly: $ Tr (M^{1} dM)^k$, k =1,3 . One can check that these are closed and invariant forms, which by some general reasons is enough to ensure that they are generators of the cohomology.. Again, I hope you would agree that these are very explicit coordinatewritten expressions which "everyone can understand". Notation Let us give similar description for invariant vector fields and get in contact with beautiful story of Capelli identity. Let introduce vector fields $\partial_{x},\partial_{y}, \partial_{z},\partial_{u}$ and form a matrix: $$ \partial_M=\begin{bmatrix} \partial_x & \partial_y \\\\ \partial_z & \partial_u \end{bmatrix} $$ Lemma 3 (Invariant vector fields) take any 2x2 number valued matrix "K", then $Tr(KM \partial_M^t )$  right invariant vector field on the group GL(2). $Tr(K\partial_M^t M)$  left invariant field on the group GL(2); Capelli's story, Howe duality, Langlands correspondence, Talalaev's formula The 2x2 matrix $ E = M (\partial_M)^t$ is a matrix with noncommutative elements. Nevertheless for the purpose of some Lie algebra theory one wants to calculate its determinant. Actually this can be done: Capelli identity: $ det^{column} (E +diag(1,0) ) = det(M) det(\partial_M) $ One of the points of Capelli's story is that this determinant is generator of the center of U(gl(2)), another point related to Howe's duality is that we can express it via left or right invariant vector fields and will get essentially the same result (up to details). It is related to the fact that GL(n) acting from the left on Mat_n is Howe's dual to the GL(n) acting from the right  and hence images of the centers for the two action must coincide. There are some modern developments around this. Let us introduce auxiliary variable "z" and define a new matrix: $$ T= (Id \partial_z E/z)$$ the point is that it will be Manin matrix  class of matrices which are "slighly noncommutative", but behave like commutative ones. And toy model for Talalaev's theorem states $$det(T)$$ will generate the center of U(gl(2)). "Nontoy" is similar result for loop algebra for gl(n). Moreover the differential operator in "z" is related to the socalled GL(n)oper which is Langlands dual to the corresponding irrep of loop algebra. Sorry for selfadvertisement: one may look at http://arxiv.org/abs/0711.2236 "Manin matrices and Talalaev's formula" for further info. 

