Conversation started Apr 14, 2013 at 11:43.
Gustavo Bandeira
Gustavo Bandeira
Apr 14, 2013 11:43
@Nimza What's the meaning of $\smile$ in $f^\ast (a \smile b) = f^\ast(a) \smile f^\ast(b)$?
Nimza
Nimza
@robjohn hi, do you know some upper estimates on grow of $\pi$ function $\pi(z) = \frac{1}{\Gamma(z+1)}$ along vertical lines?
robjohn
robjohn
@Nimza There is a question about that... just let me look
Nimza
Nimza
@GustavoBandeira it is cup product, do you know what is it for $a \in H^{p}$ and $b \in H^q$?
Gustavo Bandeira
Gustavo Bandeira
@Nimza No.
Nimza
Nimza
@GustavoBandeira okay, let $x$ be a singular ($p+q$)-simplex in topological space $X$, i.e. continuous map $x \colon \Delta^{p+q} \to X$, where $\Delta^{p+q}$ is a standard $(p+q)$-simplex
robjohn
robjohn
Apr 14, 2013 11:47
@Nimza Take a look at this answer
Nimza
Nimza
@GustavoBandeira then we can define operation of taking front $p$-face and rear $q$-face of it
@robjohn thank you!
@GustavoBandeira Then $(a \smile b) (x) = a(front\;face\;of\;x)b(rear\;face\;of\;x)$, but our homology groups need to be computed over the ring to be available to multiply here
so $a \smile b \in C^{p+q}$ if $a \in C^p$ and $b \in C^q$
@GustavoBandeira sorry, I told you about cup product in cochains, but it descends on cohomology groups obviously
Gustavo Bandeira
Gustavo Bandeira
@Nimza But I'm not understanding what you're saying.
I'm still a noob.
Nimza
Nimza
@GustavoBandeira on which stage? I'm noob too but this thing seems very easy
Gustavo Bandeira
Gustavo Bandeira
@Nimza Starting analysis.
Nimza
Nimza
aa
@GustavoBandeira we can define it on usual simplexes, if $a$ if a function that takes a $p$-simplex as input and returns for example a real number and $b$ is the similar function that takes $q$ simplex as input then $a \smile b$ is a function that takes $(p+q)$-simplex as input as returns a real number that is $a$(front $p$-face)$\cdot b$(rear $q$ face), is it clear?)
 
Conversation ended Apr 14, 2013 at 11:57.