# Prove the $R$-module isomorphism $P\oplus P\cong R\oplus R$

Let $$R=\{f:\mathbb{R}\to\mathbb{R}:f\text{ is continuous and }f(x+\pi)=f(x)\}$$ $$P=\{f:\mathbb{R}\to\mathbb{R}:f\text{ is continuous and }f(x+\pi)=-f(x)\}$$

Then under addtition and multiplication $$R$$ is a ring and $$P$$ is an $$R$$-module. Show that there is an $$R$$-module isomorphism

$$P\oplus P\cong R\oplus R$$

Background: We are learning homonological algebra (just getting started) at the moment, so maybe something about it should be made use of.

My attemp: I have tried to construct some suitable mapping $$R\to P$$ or $$P\to R$$. For example I have tried to "flip" the graph of $$f\in R$$ in $$[(2k-1)\pi,(2k)\pi]$$ to obtain a function in $$P$$. But I cannot seem to find the right one.

Questions:

(1) Am I suppose to construct this way, or should I use some magical property of $$P$$ and $$R$$?

(2) Why does the direct sum appear here; what is so important or necessary about it?

• Perhaps the way to prove the claim that's intended would be to show that there exist $(p,p')$ in $P\oplus P$ such that for any $(p_1,p_2)$ in $P\oplus P$, there are $(r_1,r_2)$ in $R\oplus R$ such that $(p_1,p_2) = (r_1.p,r_2.p')$ and that the choice of $r_1,r_2$ is unique. – Rylee Lyman May 30 at 10:39
• @RyleeLyman I don't think so... If what you say is true, then restricting this mapping to one component yields an isomorphism $P\cong R$. And the direct sum would be redundant. – trisct May 30 at 11:14

You can use the matrix $$M=\begin{pmatrix}\cos&-\sin \\ \sin&\cos\end{pmatrix}$$. Note that if $$f\in P$$, then $$f\times \sin\in R$$, similarly $$f\times \cos\in R$$. Thus if $$\begin{pmatrix}f\\g\end{pmatrix}\in P\oplus P$$ then $$M\begin{pmatrix}f\\g\end{pmatrix}= \begin{pmatrix}f\times \cos-g\times\sin\\ f\times\sin+g\times\cos\end{pmatrix} \in R\oplus R$$. Now $$M$$ is invertible, in fact its inverse is $${}^tM$$. This gives the isomorphism $$P\oplus P\simeq R\oplus R$$.

Edit : I realize that I did not really answer your questions, and specifically let me say a few words about (2). Note that a map $$R\to P$$ is entirely determined by the image of $$1$$ which is thus an element of $$P$$. Say this image is $$u$$ then the map $$R\to P$$ is given by $$f\mapsto uf$$. Now in this particular case, this cannot be an isomorphism. Indeed, by continuity $$u$$ has to vanish, so if $$f\in\operatorname{Im}(R\to P)$$ then $$f$$ has to vanish at the zeroes of $$u$$. But since there are functions in $$P$$ with no common zeroes, so $$R\to P$$ cannot be onto.

Similarly, a map $$R\oplus R\to P\oplus P$$ is also given by a $$2\times 2$$ matrix as above. This times it works because the determinant of this matrix does not vanish.

There is also a geometrical interpretation : $$R$$ is the set of continuous function on the circle whereas $$P$$ is the set of sections of the twisted line bundle $$L$$. The point is that $$L\oplus L$$ is the trivial bundle.