# Calculate homotopy groups of $\mathbb{Z}_2$-equivariant loop spaces of "complex" topological spaces

Let $$X$$ be a topological space such that complex conjugation is defined (e.g. $$\mathbb{C}^n$$) and let us define the set of maps $$S_d:= \left\{f: (I^d,\partial I^d)\to (X,x_0)\mid \overline{f(k)} = f(-k)\right\} \\\subseteq \left\{ f: (I^d,\partial I^d)\to (X,x_0) \right\} = \Omega^dX,$$ where $$I = [-1,1]$$.

Equip the sets $$S_d$$ and $$\Omega^dX$$ with the Compact-open topology, such that they become topological spaces. What can we say about the homotopy groups $$\pi_n(S_d,c_{x_0})$$, where $$c_{x_0}$$ is the constant map into $$x_0$$?

I am looking for a strategy in computing $$\pi_n(S_d,c_{x_0})$$. What I do know are the homotopy groups $$\pi_n(\Omega^dX,c_{x_0})\cong \pi_{n+d}(X,x_0)$$, which is a standard result in homotopy theory. But $$S_d$$ is a subspace in $$\Omega^dX$$, which does not have to share the same homotopy groups. The elements of $$S_d$$ satisfy a certain $$\mathbb{Z}_2$$-equivariance condition and the theory about $$G$$-equivariant homotopy seems to be very involved, although I would certainly dive into it, when I knew that there were tools with which one could calculate the homotopy groups of $$S_d$$.

Edit: Consider as an example $$X = V_p(\mathbb{C}^q)$$, the complex Stiefel manifold, whose elements we will interpret as complex $$q\times p$$-matrices.

Edit 2: The set of path components $$\pi_0(S_d,c_{x_0})$$ would be sufficient.

Edit 3: Question has been answered (see below).

• Maybe this article could help. Notice that your space $X$ is a $\mathbb{Z} / 2 \mathbb{Z}$-space under the conjugation. Commented Apr 10, 2023 at 12:45

$$\newcommand{\Z}{\mathbf{Z}}\newcommand{\Map}{\mathrm{Map}}$$Let $$\sigma$$ denote the sign representation of $$\Z/2$$, and let $$S^{d\sigma}$$ denote the one-point compactification of $$\sigma^{\oplus d}$$. Let $$X$$ be a space with a $$\Z/2$$-action. It seems that you're interested in $$(\Omega^{d\sigma} X)^{\Z/2}$$ (this is your $$S_d$$). If $$[d-1] = \{0,\cdots,d-1\}$$, this can be described as the total homotopy fiber of the functor $$2^{[d-1]} \to \mathrm{Spaces}_\ast$$ sending $$J \mapsto \Map_{\Z/2}((\Z/2)^J_+, X)$$. This comes from the $$\Z/2$$-equivariant cell structure of $$S^{d\sigma}$$: it is the total homotopy cofiber of the functor $$(2^{[d-1]})^\mathrm{op} \to \mathrm{Spaces}_\ast$$ sending $$J\mapsto (\Z/2)^J_+$$.
For instance, when $$d = 1$$ (you can induct on $$d$$ to get the general claim from this case), recall that $$S^\sigma$$ can be constructed as the homotopy cofiber of the crushing map $$(\Z/2)_+ \to S^0$$. This implies that there's a fiber sequence $$(\Omega^\sigma X)^{\Z/2} = \Map_{\Z/2}(S^\sigma, X) \to \Map_{\Z/2}(S^0, X) = X^{\Z/2} \to \Map_{\Z/2}((\Z/2)_+, X) = X,$$ the second map being the inclusion of the fixed points. This is precisely what it means for $$(\Omega^\sigma X)^{\Z/2}$$ to be the total homotopy fiber of the functor $$2^{[0]} = \Delta^1 \to \mathrm{Spaces}_\ast$$. Note that this fiber sequence gives a long exact sequence on homotopy groups $$\cdots \to \pi_{j+1}(X) \to \pi_j((\Omega^\sigma X)^{\Z/2}) \to \pi_j(X^{\Z/2}) \to \pi_j(X) \to \cdots$$
For example, if $$X = \mathrm{BU}(n)$$ equipped with complex conjugation, this is going to imply that $$(\Omega^\sigma \mathrm{BU}(n))^{\Z/2}$$ is the homotopy fiber of the map $$\mathrm{BO}(n) \to \mathrm{BU}(n)$$, i.e., is $$\mathrm{U}(n)/\mathrm{O}(n)$$. In your case, if one equips $$V_m(\mathbf{C}^n) = \mathrm{U}(n)/\mathrm{U}(n-m)$$ with the $$\Z/2$$-action given by complex conjugation, this is going to imply that $$(\Omega^\sigma V_m(\mathbf{C}^n))^{\Z/2}$$ is the total homotopy fiber of the square $$\require{AMScd} \begin{CD} \mathrm{BO}(n-m) @>>> \mathrm{BU}(n-m)\\ @VVV @VVV \\ \mathrm{BO}(n) @>>> \mathrm{BU}(n). \end{CD}$$ Not sure if this has a simpler description.
Indeed, $$\pi_{D}\left(\left(\Omega^{d+1}X\right)^{\mathbb{Z}_2},c_{x_0}\right)\cong\pi_{D+1}\left(\Omega^{d}X,\left(\Omega^{d}X\right)^{\mathbb{Z}_2},c_{x_0}\right)$$ for any $$\mathbb{Z}_2$$-space $$X$$ with $$\mathbb{Z}_2$$-fixed point $$x_0\in X^{\mathbb{Z}_2}$$ which gives rise to the long exact sequence of the pair $$\left(\Omega^{d}X,\left(\Omega^{d}X\right)^{\mathbb{Z}_2}\right)$$. The map $$c_{x_0}$$ denotes the constant map to $$x_0$$.