I bumped into the equvariant cohomology while reading about torus action.
I'm trying to understand the concept of formality, that is, for a space $X$ with a torus action of $T$, it is formal if $H_T^*(X) $ is free as a module over $H^*(BT)$ . Let's assume field coefficient for simplicity.
I have found a lot of studies determining when a space is formal in this sense, and many authors mention that some symplectic manifold, toric varieties, etc fall into this category.
However, I haven't seen an explicit example of non-formal $T$- spaces . I think that if the action is free, then the space won't be formal (am I right?), however I would like to know any other example of a Manifold/Finite CW-complex out there where the action is not free and the space is not formal.
Take $T=U(1)\times U(1)$ as a maximal torus in $U(2)$ by the embedding into the diagonal matrices. This gives a free (left) action of $T$ on $U(2)$. We want to calculate the equivariant cohomology of $U(2)$ with respect to this torus action, and this is the ordinary cohomology of the space $ET\times_TU(2)$, where $ET$ is a contractible free right $T$-space. That is, $H^*_T(U(2))=H^*(ET\times_TU(2))$.
Observe that there is a fibration $U(2)\xrightarrow U(2)/(U(1)\times U(1))\cong S^2$, and this induces a fibration sequence
$ET\rightarrow ET\times_TU(2)\xrightarrow{p} U(2)/(U(1)\times U(1))$.
Since $ET$ is contractible it follows that the map $p$ is a homotopy equivalence, so that $ET\times_TU(2)\simeq S^2$. In particular
$H^*_T(U(2))=H^*(ET\times_TU(2))\cong H^*S^2$
and therefore $U(1)$ cannot be free over the polynmial algebra $H^*T\cong \mathbb{Z}[x_2,y_2]$. This gives an example of a non-formal free $T$-space.
In fact if you examine the Serre spectral sequence of $U(2)\rightarrow ET\times_TU(2)\xrightarrow{q}BT$ you find that you can choose a generator $a_1\in H^1U(2)\cong\mathbb{Z}$ that transgresses to the class $x_2-y_2\in H^2BT$. Then $s_2:=q^*x_2=q^*y_2$ generates $H^2(S^2)\cong \mathbb{Z}$. Thus the $H^*BT$-module structure of $H^*_T(U(2))\cong\Lambda(s_2)$ is given by
$x_2\cdot 1=y_2\cdot 1=s_2$,
$x_2\cdot s_2=y_2\cdot s_2=0$.