maths researcher

or perhaps retired

any math profs here?

@TedShifrin

yupp

@TedShifrin No. I want to use Mayer vietoris. My decomposition of $\mathbb{S}^1\vee \mathbb{S}^1$ is $\mathbb{S}^1\cup \mathbb{S}^1$ such that $\mathbb{S}^1\cap \mathbb{S}^1$ is the wedged point.

I am identifying $\mathbb{S}^1$ with its image under canonical map $\mathbb{S}^1\rightarrow \mathbb{S}^1\vee \mathbb{S}^1$

because, $\mathbb{S}^1\vee \mathbb{S}^1 = \mathbb{S}^1\cup \mathbb{S}^1$ where $\mathbb{S}_1 \cap \mathbb{S}^1$ is a point

Could one do that $H_k(\mathbb{S}^1\vee \mathbb{S}^1)=H_k(\mathbb{S}^1)\oplus H_k(\mathbb{S}^1)$ without appealing to good pairs

Oh, I thought you meant for me to try to find a theorem that would allow me to deduce such a conclusion

that would mean that $\mathbb{R}^2$ is isometric to $\mathbb{R}^2$ with standard metric or the upper half plane

Is the theorem Killing-Hopf theorem @TedShifrin

So all Riemannian metrics on $\mathbb{R}^2$ (compatible with the standard topology) give sectional curvature 0?

I would say no, but i'm not sure of any theorem that would allow me to deduce such a conclusion

Does $\mathbb{R}^2$ admit a complete metric with sectional curvate $k\geq 1$?

Can I modify the product metric on the cylinder to give me a riemannian metric which preserves completeness but now the cylinder has sectional curvature atleast 1?

Let $p(t)=e^{2\pi it}$ then the covering map is $p\times id_{\mathbb{R}}$ I assume

We haven't discussed quotient spaces and manifold structures on quotient spaces

Euclidean metric, ofcourse @TedShifrin

My notes say the following: There exists a smooth covering map $\pi: \mathbb{R}^2\rightarrow \mathbb{S}^1\times \mathbb{R}$. Therefore, $\mathbb{S}^1\times \mathbb{R}$ has a metric for which $\pi$ is a local isometry and under such a metric, $\mathbb{S}^1\times \mathbb{R}$ is complete

Yes, but i'm not looking at product metric @TedShifrin

So consider $\mathbb{S}^1\times \mathbb{R}$. Since there exists a smooth covering map $\pi: \mathbb{R}^2\rightarrow \mathbb{S}^1\times \mathbb{R}$, $\mathbb{S}^1\times \mathbb{R}$ admits a Riemanian metric making it into a complete Riemannian manifold under $\pi^{*}g$, where $g$ is euclidean metric. here, $\pi$ is also a local isometry, and so this means under this metric $\mathbb{S}^1\times \mathbb{R}$ has constant sectional curvature 0, right?

Yeah, I assume you would need zorn's lemma or something of that sort

the extension may be uncountable, so i'm not assuming separability or that it has a schauder basis

If $(e_n)_n$ is an orthonormal sequence in a hilbert space $H$, can we extend it to a basis to all of $H$?

if $M$ is a Riemannian manifold and $p\in M$ then for any two linearly independent vectors $u,v\in T_pM$, $sec(u,v)=\frac{\langle R(u,v)v,u\rangle}{|u|^2|v|^2-\langle u,v\rangle^2}$ and $sec(u,v)$ is independent of the vectors you choose in $span(u,v)$

isn't sectional curvature independent of the vectors you choose?

why doesn't my argument show that $sec(u,v)\geq 0$ for any two linearly indepoendent vectors $u,v$?

yes, that I know. There's something I don't understand or confused by, but not sure what exactly

that's my issue, i'm trying to see why what i'm doing doesn't make sense

which bit?

for which bit? Denominator $1$ because $|u|^2=|v|^2=1$ and $\langle u,v\rangle =0$ and inner product always non negative, no?

So that would mean the denominator is just $1$ and $sec(u,v)\geq 0$

$sec(u,v)= \frac{\langle R(u,v)v,u\rangle}{|u|^2|v|^2- \langle u,v \rangle^2}$ for $u,v$ linearly independent tangent vectors. Can't we always assume $u,v$ are orthonormal?

what three cases?

I don't understand, can't we just pick an orthonormal basis in our product tangent space and then we would get $sec(u,v)=\langle R(u,v)v , u \rangle$ which is always non nehative?

then $M_1\times M_2$ with the product metric has non negative sectional curvature as well, right?

If $M_1$ and $M_2$ have non negative sectional curvature

Hi

What do you mean?

ahh okay makes sense

our rings have identity

ofcourse

is its generator

1

oh

o

In this case $H^m(\mathbb{S}^n;\mathbb{Z})$ has a generator for each $m$

yes 0 is a generator

if $R=\mathbb{Z}$ then only $H^m(\mathbb{S}^n,\mathbb{Z})=\mathbb{Z}$ if $m=n$ and $0$ otherwise

@TedShifrin Isn't that what the proof Lemma 57.11 is saying in uni-regensburg.de/Fakultaeten/nat_Fak_I/friedl/… ?