Homology class of sphere in projective space

Question

Let $\mathbb{F}_2$ be the field with two elements. Then consider the homology of the real projective space with $\mathbb{F}_2$ coefficients $H_k(\mathbb{RP}^n; \mathbb{F}_2)\cong\mathbb{F}_2$ for $0< k< n$. I wonder which homology classes can be realized by an embedded sphere $S^k\subset\mathbb{RP}^n$. Clearly, if we embed $S^k$ as a compact sphere to some affine part $\mathbb{R}^n$ we realize the trivial class. For $k=1$ we can just take a projective line in $\mathbb{RP}^n$ to get the nontrivial class in $H_1(\mathbb{RP}^n; \mathbb{F}_2)$. But can we realize the nontrivial class in $H_k(\mathbb{RP}^n; \mathbb{F}_2)$ by a sphere for $1<k<n$?

Answer 1

The answer is no for $k>1$ : suppose you have a map $S^k\to \mathbb RP^n$, with $k>1$. Then $S^k$ is simply-connected, so this map lifts to $S^k\to S^n\to \mathbb RP^n$ (where $S^n\to \mathbb RP^n$ is the standard covering map).

It follows that for $k<n$, the map on homology factors as $H_k(S^k)\to 0 \to H_k(\mathbb RP^n)$

For $k=n$, one has to note that $H_n(S^n)\to H_n(\mathbb RP^n)$ is $0$ with $\mathbb F_2$-coefficients (to prove that, one should differentiate between the odd case and the even case : in the even case, $\mathbb RP^n$ is not orientable, so its integral homology vanishes so it's easy; in the odd case, in integral homology, the map is multiplication by $2$ by the local degree formula)