Is there any reference for calculation of the rational homology of the free loop space $H_*(\mathcal{L}Gr(k,n),\mathbb{Q})$ of a complex Grassmanian? More precisely, I am interested in computing ranks $$r_i=rk(H_i(\mathcal{L}Gr(2,4),\mathbb{Q}))$$ (Here $Gr(k,n)$ stands for kplanes in nspace).
The complex Grassmannian $Gr(2,4)$ can be realized up to homotopy as the homotopy fiber of the map $BU(2) \times BU(2) \rightarrow BU(4)$ which corresponds to the Whitney sum of two complex rank 2 bundles. To obtain the rational cohomology of it (and then of its free loop space), we'll calculate its minimal model as the cofiber of the map this topological map induces on minimal models. This lengthy and perhaps unnecessarily detailed calculation ends with "end of calculation" below; so skip there to just get the minimal model.
The minimal model (in the sense of rational homotopy theory) of $BU(2)$ is given by $\Lambda(a_1,a_2)$, and so the model of $BU(2)\times BU(2)$ is given by $\Lambda(a_1,a_2)\otimes \Lambda(b_1,b_2) = \Lambda(a_1,a_2,b_1,b_2)$. Let us denote the minimal model of $BU(4)$ by $\Lambda(c_1,c_2,c_3,c_4).$ The differential is zero in all of these algebras, and the variables $a_i,b_i,c_i$ are of degree equal to twice their index (they are universal Chern classes). The notation $\Lambda$ is used for the free exterior (gradedcommutative) algebra on a given set of generators in prescribed degrees.
On the level of cohomology (or equivalently, minimal models, since everything is formal), the map $BU(2)\times BU(2) \rightarrow BU(4)$ is given by $H^*(BU(4)) \rightarrow H^*(BU(2))\otimes H^*(BU(2))$, that is, $$\Lambda(c_1,c_2,c_3,c_4) \rightarrow \Lambda(a_1,a_2,b_1,b_2).$$
To figure out the image of $c_i$, observe that if a complex rank 4 bundle $E$ splits into the sum of two complex rank 2 bundles, $C = A\oplus B$, then on Chern classes we have $$1+c_1(C) + c_2(C) + c_3(C) + c_4(C) = (1+c_1(A)+c_2(A))(1+c_1(B)+c_2(B)).$$ Denoting by $a_i,b_i,c_i$ the $i$th Chern class of $A$,$B$,or $C$ respectively, we obtain the equations \begin{align*} c_1 &= a_1+b_1, \\ c_2 &= a_2+b_2+a_1b_1, \\ c_3 &= a_1b_2 + a_2b_1, \\ c_4 &= a_2b_2. \end{align*} These relations tell us how the map of minimal models $\Lambda(c_1,c_2,c_3,c_4) \rightarrow \Lambda(a_1,a_2,b_1,b_2)$ is prescribed.
Now we turn this map $\Lambda(c_1,c_2,c_3,c_4) \rightarrow \Lambda(a_1,a_2,b_1,b_2)$ into a fibration. That is, we will find a differential graded algebra $E$ containing $\Lambda(c_1,c_2,c_3,c_4)$ with a quasiisomorphism to $\Lambda(a_1,a_2,b_1,b_2)$ such that the map $\Lambda(c_1,c_2,c_3,c_4) \to \Lambda(a_1,a_2,b_1,b_2)$ factors into the inclusion followed by the quasiisomorphism.
This should be thought of as dual to the mapping path space construction, turning any map $X\to Y$ into a fibration. Just as the homotopy fiber is the fiber of the map from the mapping path space to $Y$, here the algebraic model of the homotopy fiber is the fiber of the inclusion $\Lambda(c_1,c_2,c_3,c_4) \hookrightarrow E$, which is defined to be $E/\textrm{ideal}(c_1,c_2,c_3,c_4)$.
Now we have to construct this $E$ to contain $\Lambda(c_1,c_2,c_3,c_4)$ and to be quasiisomorphic via some map $f$ to $\Lambda(a_1,a_2,b_1,b_2)$. In order to do this, prescribe $f(c_i)$ to be the corresponding polynomial in $a_i,b_i$ given above. Introduce a variable $\overline{a_i}$ in degree 2 so that $f(\overline{a_1}) = a_1$. Note that now we have $a_1$ and $b_1$ in the image of $f$, since we have $a_1+b_1$ thanks to $c_1$ and $a_1$ is the image of $\overline{a_1}$. Next, introduce a variable $\overline{a_2}$ such that $f(\overline{a_2}) = a_2$. This gets us $a_1,b_1,a_2,b_2$ in the image of $f$, due to $c_1$,$c_2$,$\overline{a_1}$, $\overline{a_2}$. So, we have made $f$ surjective. Now we look at its kernel, and introduce variables to kill these elements in cohomology. Since $c_3$ has to be mapped to $a_1b_2 + a_2b_1$, we see that we necessarily have $$f(c_3  \overline{a_1}c_2 + 2\overline{a_1}\overline{a_2} + \overline{a_1}^2c_1  \overline{a_1}^3 + \overline{a_2}c_1) = 0.$$ Similarly, since $f(c_4) = a_2b_2$, we get that $$f(c_4  \overline{a_2}c_2 + \overline{a_2}^2 + \overline{a_1}\overline{a_2}c_1  \overline{a_1}^2\overline{a_2}) = 0.$$ To get rid of this noninjectivity on cohomology, we introduce variables $\eta_5$ and $\eta_7$ in $E$ (in degrees 5 and 7, respectively) with the prescription that \begin{align*} d\eta_5 &= c_3  \overline{a_1}c_2 + 2\overline{a_1}\overline{a_2} + \overline{a_1}^2c_1  \overline{a_1}^3 + \overline{a_2}c_1, \\ d\eta_7 &= c_4  \overline{a_2}c_2 + \overline{a_2}^2 + \overline{a_1}\overline{a_2}c_1  \overline{a_1}^2\overline{a_2}. \end{align*} Setting $f(\eta_5) = f(\eta_7) = 0$, we have made $f$ injective as well on cohomology, and thus a quasiisomorphism.
So, our $E$ is given by $\Lambda(c_1,c_2,c_3,c_4, \overline{a_1}, \overline{a_2} , \eta_5, \eta_7)$ with the first six variables closed, and the differential on $\eta_5$ and $\eta_7$ given as above. Now we take the cofiber of this map, that is, we take $\Lambda(c_1,c_2,c_3,c_4, \overline{a_1}, \overline{a_2} , \eta_5, \eta_7)$ mod the elements in the differential ideal generated by the $c_i$. Our cofiber is given by $\Lambda(\overline{a_1}, \overline{a_2}, \eta_5, \eta_7)$, where $d\overline{a_1} = d\overline{a_2} = 0$ and \begin{align*}d\eta_5 &= 2\overline{a_1}\overline{a_2}  \overline{a_1}^3, \\ d\eta_7 &= \overline{a_2}^2  \overline{a_1}^2\overline{a_2}.\end{align*}
For notational convenience in what follows, let us rename the variables to $c_1 = \overline{a_1}, c_2 = \overline{a_2}, u = \eta_5, v = \eta_7$.
END OF CALCULATION
So we have that the minimal model of $Gr(2,4)$ is given by $$\Lambda(c_1,c_2, u,v), \textrm{ where }dc_1 = dc_2 = 0, du = c_1(c_1^2  2c_2), dv = c_2^2  c_1^2c_2.$$
As for the free loop space, we use the first algebraic construction in Section 11 of Sullivan's Infinitesimal Computation in Topology. To form the minimal model of $LGr(2,4)$, we take the minimal model of $Gr(2,4)$, introduce a new copy of all the generators but with degree shifted down by one, define a derivation $i$ on this new free algebra given by declaring it sends a generator from the original minimal model to its shifteddownbyone counterpart, and sends these new shifted generators to 0 (and extending so that it is a derivation). The differential on this new larger algebra extends the old differential and satisfies $di+id=0$. Concretely, in our case, the minimal model of $LGr(2,4)$ will have underlying algebra $$\Lambda(c_1,c_2, u,v, ic_1, ic_2, iu, iv),$$ where $deg(ic_1) = 1, deg(ic_2) = 3, deg(iu) = 4, deg(iv) = 6$. The condition $di+id = 0$ gives us $d(ic_1) = 0$ and $d(ic_2) = 0$, along with $d(iu) = (ic_1)\cdot (c_1^22c_2) + c_1 \cdot (2c_1\cdot ic_1  2ic_2)$ and $d(iv) = 2c_2 \cdot ic_2 + 2c_1 \cdot ic_1 \cdot c_2 + c_1^2 \cdot ic_2$.
From here we can now calculate the ranks of the rational (co)homology groups, certainly with a computer algebra system, and maybe with a clever closedform expression (though I don't know it, sorry). The first few ranks are $r_1 = 1, r_2 = 1, r_3 = 2, r_4 = 3, r_5 = 2$. By a theorem of Sullivan and ViguéPoirrier, this sequence of ranks will be unbounded.
A similar calculation will give you the rational homology of the free loop space of other Grassmannians; just consider $BU(k) \times BU(nk) \rightarrow BU(n)$ instead.

10$\begingroup$ A faster route to the minimal model is to note that the Grassmannian is a smooth compact variety, hence formal. So it suffices to resolve its cohomology ring. But its cohomology ring is a complete intersection (see e.g 3264 and all that sec 5 for the regular sequence). Thus a free resolution is given by a Koszul complex, as in the outcome of your computation. $\endgroup$ – Phil Tosteson Apr 5 at 4:52

$\begingroup$ @Aleksandar Thanks a lot! I have just left puzzled with the degree 5 rank: shouldn't it be generated by $c_1 ic_2, c_2 ic_1, c_1^2 ic_1$ but having a relation between them coming from $d(iu)$ should kill the rank by 1, hence $r_5=2$ ? $\endgroup$ – Filip92 Apr 8 at 14:50

$\begingroup$ @Filip92 Ah right, thanks, I was hasty. I’ve edited accordingly. $\endgroup$ – Aleksandar Milivojevic Apr 8 at 15:13