is there a ring homomorphism $f:H^*(T^n;R)\to H^*(T^n;R)$ of graded rings which is not induced by a continuous map $T^n\to T^n$?

Question

Let $R$ be a commutative ring with unit $1_R$ and let $\xi\in H^1(S^1;R)$ be a generator ($H^1(S^1;R)$ is the first singular cohomology group of $S^1$). Let $p_i:(S^1)^n\to S^1$ be the projection onto the i-th coordinate and $H^1(p_i):H^1(S^1;R)\to H^1((S^1)^n;R)$ it's induced map on singular cohomology. We set $H^1(p_i)(\xi)=:\xi_i$, notice that $(S^1)^n=T^n$ is the n-torus. The cohomology ring of $T^n$, $H^*(T^n;R)$, is isomorphic to the exterior algebra over $R$, $\Lambda _R[\xi_1,..,\xi_n]$ (as graded rings).

My question is: is there a ring homomorphism $f:H^*(T^n;R)\to H^*(T^n;R)$ of graded rings which is not induced by a continuous map $T^n\to T^n$, i.e. does not come from a continuous map $T^n\to T^n$?

Maybe it is sufficient to find out, if every multiplicative map $H^1(T^n;R)\to H^1(T^n;R)$ comes from a continuous map $T^n\to T^n$, because a ring homomorphism of graded rings preserves the grading. But Then I have no idea how to continue. Do you have an idea?

Answer 1

It depends on the ring. Consider $R = \mathbb{Q}$, for example. Then there is no continuous map $S^1 \to S^1$ inducing the graded ring morphism $\Lambda_\mathbb{Q}(\xi_1) \to \Lambda_\mathbb{Q}(\xi_1)$ sending $\xi_1$ to $\frac{1}{2} \xi_1$. Indeed, every map $S^1 \to S^1$ is homotopic to one of the maps $z \mapsto z^n$ for some $n \in \mathbb{Z}$, and the map it induces on cohomology sends $\xi_1$ to $n \xi_1$.

But it's true if $R = \mathbb{Z}$. Let $f : H^*(T^n) \to H^*(T^n)$ be some graded ring morphism. Then $f$ is uniquely determined by the images of what you call $\xi_1, \dots, \xi_n$. Since $f(\xi_i) \in H^1(T^n)$, one has: $$f(\xi_i) = \sum_{j=1}^n a_{i,j} \xi_j$$ for some $a_{i,j} \in \mathbb{Z}$.

Now define $\varphi : T^n \to T^n = (S^1)^n$ to be the product of the maps $\varphi_i : T^n \to S^1$ given by: $$\varphi_i(z_1, \dots, z_n) = z_1^{a_{1,i}} \dots z_n^{a_{n,i}}.$$

Then you can check that $\varphi^*(\xi_i) = \varphi^*(p_i^*(\xi)) = \varphi_i^*(\xi)$ is indeed given by $\sum_j a_{i,j} \xi_j$.

Answer 2

When $R=\Bbb Z$ the answer to your question is no:

Since $T^n$ is a $K(\pi_1(T^n),1)$, every group endomorphism of $\pi_1(T^n)$ is induced by a map $T^n\to T^n$. (See for example Prop 1B.9 in Hatcher). Therefore, any endomorphism of $\pi_1(T^n)^{ab}=H_1(T^n)=H^1(T^n)$ is also induced by a map of spaces.

Since $H^*(T^n)\cong \bigwedge_{\Bbb Z}\left[ \alpha_1,\cdots,\alpha_n \right]$ with $\deg \alpha_i=1$, any morphism $H^*(T^n)\to H^*(T^n)$ is determined by what it does in degree $1$, hence it is induced by a map of spaces.