Given a continuous map $f:S^n \to S^n$, the homological degree of $f$ is defined to be the integer $\deg f_*$ such that $f_*(\alpha)=(\deg f_*)\alpha$ for any $\alpha \in H_n(S^n) \approx \mathbb{Z}$.

Exercise 2.2.8 in Hatcher's text asks the reader to consider a complex polynomial $p:\mathbb{C} \to \mathbb{C}$. If this polynomial is nonconstant, then setting $p(\infty)=\infty$ gives an extension of $p$ to the one point compactification $p:\hat{\mathbb{C}} \to \hat{\mathbb{C}}$. But $\hat{\mathbb C} \approx S^2$, so one can consider the homological degree of $p$. Turns out, this degree is equal to the degree of $p$ as a polynomial. $$\deg p_* = \deg p$$

If you do this whole thing over $\mathbb{R}$ instead, with one point compacitication $\hat{\mathbb R} \approx S^1$, you don't get an analagous result. If $p(x)=a_nx^n + \dotsb +a_1x+a_0$ with $a_i \in \mathbb{R}$ (and $a_n \not= 0$), then $$\deg p_* = \begin{cases} 0 & \text{if $n$ is even} \\ \operatorname{sign} a_n & \text{if $n$ is odd} \end{cases}$$ where $\operatorname{sign}$ is the $\pm 1$ sign of a nonzero real number.

This led me to wonder what happens in other scenarios. For instance, what happens over the quaternions $\mathbb{H}$ with $\hat{\mathbb H} \approx S^4$? I'm not even sure what "quaternionic polynomial" would mean since the quaternions aren't commutative, so expressions like $i \cdot q \cdot i$ and $(i^2)q=-q$ aren't equivalent.

What about other complete fields, like $p$-adics $\mathbb{Q}_p$? These spaces are much worse topologically, in particular totally disconnected. But one can still consider the projective lines $\mathbb{P}^1(\mathbb{Q}_p)$ (which is the same as the one point compactification in the $\mathbb{R}$ and $\mathbb{C}$ case at least). I imagine that number theorists and algebraic geometers have invented some appropriate analog of singular homology to study maps of these things.

  • $\begingroup$ It seems that for the quaternions one usually only considers one-sided polynomials. See core.ac.uk/download/pdf/132798758.pdf. $\endgroup$ Nov 10, 2021 at 11:35
  • $\begingroup$ If you adopt the scheme-theoretic viewpoint, then it's always true that the degree of the map $\Bbb P^1_k\to\Bbb P^1_k$ given by $[f(x):1]$ is $\deg f$ for any field $k$. $\endgroup$
    – KReiser
    Nov 15, 2021 at 5:37
  • $\begingroup$ @KReiser can you say a bit more what you mean by that? I really know very little about algebraic geometry. $\endgroup$ Nov 15, 2021 at 5:59

1 Answer 1


For arbitrary fields, we have the étale cohomology, which is more well-behaved for algebraically closed fields.

Let $k$ be an algebraically closed field and $n$ a positive integer that is not zero in $k$. Then we consider the group $H^2(\Bbb P^1_k, \mu_n) \cong \Bbb Z/n\Bbb Z$, where $\mu_n$ is the group of $n$th roots of unity, which is non-canonically isomorphic to $\Bbb Z/n\Bbb Z$.

Now let $f: \Bbb P^1_k \to \Bbb P^1_k$ be the map induced by a polynomial $p = \sum a_n x^n$. From the computation of $H^2(\Bbb P^1_k, \mu_n)$, we have the following diagram:

$\require{AMScd}$ $$\begin{CD} \operatorname{Pic}(\Bbb P^1_k) @>>> \operatorname{Pic}(\Bbb P^1_k) @>>> H^2(\operatorname{Pic}(\Bbb P^1_k), \mu_n) @>>> 0 \\ @VgVV @VgVV @Vf^\ast VV @VVV \\ \operatorname{Pic}(\Bbb P^1_k) @>>> \operatorname{Pic}(\Bbb P^1_k) @>>> H^2(\operatorname{Pic}(\Bbb P^1_k), \mu_n) @>>> 0 \end{CD}$$

where $\operatorname{Pic}(\Bbb P^1_k) \cong \Bbb Z$ denotes the Picard group.

where $g$ sends the point $[z]$ to $p^{-1}([z])$, so $g$ is just multiplication by $\deg p$.

Therefore, $f^\ast$ is also multiplication by $\deg p$.

Now if $k$ is not algebraically closed, we can use the Hochschild--Serre spectral sequence: $$H^r(\operatorname{Gal}(\overline{k}/k), H^s(\Bbb P^1_{\overline k}, \mu_n)) \implies H^{r+s}(\Bbb P^1_k, \mu_n)$$

Note that $\mu_n$ is no longer isomorphic to $\Bbb Z/n\Bbb Z$, because we now consider them as $\operatorname{Gal}(\overline{k}/k)$-modules.

Now $H^0(\Bbb P^1_{\overline k}, \mu_n) = \mu_n$ and $H^2(\Bbb P^1_{\overline k}, \mu_n) = \Bbb Z/n\Bbb Z$ (these are the only two non-trivial groups among $H^s(\Bbb P^1_{\overline k}, \mu_n)$), and: $$H^r(\operatorname{Gal}(\overline{k}/k),\mu_n) = \begin{cases} \mu_n \cap k & r = 0 \\ k^\times / k^{\times n} & r = 1 \\ 0 & r \ge 2 \end{cases}$$

Therefore, $$H^r(\Bbb P^1_k,\mu_n) = \begin{cases} \mu_n \cap k & r = 0 \\ k^\times / k^{\times n} & r = 1 \\ H^{r-2}(\operatorname{Gal}(\overline{k}/k), \Bbb Z/n\Bbb Z) & r \ge 2 \end{cases}$$

So $H^2(\Bbb P^1_k,\mu_n)$ is still $\Bbb Z/n\Bbb Z$, and we still recover $f^\ast$ being multiplication by $\deg p$.

  • $\begingroup$ Very nice, but only one of four rings in the question is an algebraically closed field. Remarks about any other cases would add to the answer. Can étale cohomology really not say anything over finite fields here, for instance? I have no idea, except indirectly since the whole idea of étale cohomology was afaik to support a proof of the Weil conjectures. $\endgroup$ Nov 15, 2021 at 4:28
  • $\begingroup$ @KevinArlin Done. $\endgroup$
    – Kenny Lau
    Nov 15, 2021 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.