On the definition of coboundary operator for Lie groups?












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Let $G$ be a Lie group and $p$ a positive integer. A smooth $p$-cochain on $G$ is an element of $$C^p(G):=C^infty(underbrace{Gtimes ldots times G}_{p}).$$ We can define coboundary operators $delta^p:C^p(G)longrightarrow C^{p+1}(G)$ setting $$(delta^p f)(g_0, ldots, g_p):=f(g_1, ldots, g_p)+sum_{j=1}^{p} (-1)^j f(g_0, ldots, g_{p-1} g_p, ldots, g_p)+(-1)^{p+1} f(g_0, ldots, g_{p-1}).$$ This will give us $mathbb R$-linear maps such that $delta^{p+1}circ delta^p=0$ for every $p$. Therefore we have a complex of $mathbb R$-vector spaces ${(C^p(G), delta^p)}$ and we get a cohomology theory for $G$.



What is the motivation behind the definition of those $delta^p$? Does anyone know its origins? References are also welcome.



I'm using here cohomology of Lie groups just to illustrate the situation but several others cohomologies are obtained using coboundary operators like the above one.



Thanks.










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$endgroup$

















    4












    $begingroup$


    Let $G$ be a Lie group and $p$ a positive integer. A smooth $p$-cochain on $G$ is an element of $$C^p(G):=C^infty(underbrace{Gtimes ldots times G}_{p}).$$ We can define coboundary operators $delta^p:C^p(G)longrightarrow C^{p+1}(G)$ setting $$(delta^p f)(g_0, ldots, g_p):=f(g_1, ldots, g_p)+sum_{j=1}^{p} (-1)^j f(g_0, ldots, g_{p-1} g_p, ldots, g_p)+(-1)^{p+1} f(g_0, ldots, g_{p-1}).$$ This will give us $mathbb R$-linear maps such that $delta^{p+1}circ delta^p=0$ for every $p$. Therefore we have a complex of $mathbb R$-vector spaces ${(C^p(G), delta^p)}$ and we get a cohomology theory for $G$.



    What is the motivation behind the definition of those $delta^p$? Does anyone know its origins? References are also welcome.



    I'm using here cohomology of Lie groups just to illustrate the situation but several others cohomologies are obtained using coboundary operators like the above one.



    Thanks.










    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Let $G$ be a Lie group and $p$ a positive integer. A smooth $p$-cochain on $G$ is an element of $$C^p(G):=C^infty(underbrace{Gtimes ldots times G}_{p}).$$ We can define coboundary operators $delta^p:C^p(G)longrightarrow C^{p+1}(G)$ setting $$(delta^p f)(g_0, ldots, g_p):=f(g_1, ldots, g_p)+sum_{j=1}^{p} (-1)^j f(g_0, ldots, g_{p-1} g_p, ldots, g_p)+(-1)^{p+1} f(g_0, ldots, g_{p-1}).$$ This will give us $mathbb R$-linear maps such that $delta^{p+1}circ delta^p=0$ for every $p$. Therefore we have a complex of $mathbb R$-vector spaces ${(C^p(G), delta^p)}$ and we get a cohomology theory for $G$.



      What is the motivation behind the definition of those $delta^p$? Does anyone know its origins? References are also welcome.



      I'm using here cohomology of Lie groups just to illustrate the situation but several others cohomologies are obtained using coboundary operators like the above one.



      Thanks.










      share|cite|improve this question









      $endgroup$




      Let $G$ be a Lie group and $p$ a positive integer. A smooth $p$-cochain on $G$ is an element of $$C^p(G):=C^infty(underbrace{Gtimes ldots times G}_{p}).$$ We can define coboundary operators $delta^p:C^p(G)longrightarrow C^{p+1}(G)$ setting $$(delta^p f)(g_0, ldots, g_p):=f(g_1, ldots, g_p)+sum_{j=1}^{p} (-1)^j f(g_0, ldots, g_{p-1} g_p, ldots, g_p)+(-1)^{p+1} f(g_0, ldots, g_{p-1}).$$ This will give us $mathbb R$-linear maps such that $delta^{p+1}circ delta^p=0$ for every $p$. Therefore we have a complex of $mathbb R$-vector spaces ${(C^p(G), delta^p)}$ and we get a cohomology theory for $G$.



      What is the motivation behind the definition of those $delta^p$? Does anyone know its origins? References are also welcome.



      I'm using here cohomology of Lie groups just to illustrate the situation but several others cohomologies are obtained using coboundary operators like the above one.



      Thanks.







      lie-groups homology-cohomology






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      asked Aug 19 '16 at 14:01









      PtFPtF

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          $begingroup$

          This is more an extended comment that an answer, but I believe this might give you an idea of where the inspiration comes from.



          If $G$ is not a Lie group but a Lie algebra or a $k$-algebra and $P$ is $G$-bimodule, we can similarily define $C^p(G,P)=mathrm{Hom}(Gotimescdotsotimes G, P)$. When you define the cohomology like that it becomes a derived functor (see the examples). By this I mean that if you consider the functor $Z$ sending a $G$-bimodule $P$ to $Z(P)={xin Pmid gx=xg forall gin G}$, then the family $H^n(G,-)$ are the derived functors of $Z$. Unfortunately, I'm not sure how to translate this into the case when $G$ is not an algebra, maybe if you can obtain a $G$-bimodule-like (with only the group operation) on $mathbb{R}$ everything works (in the wikipedia link there is an example of group cohomology that probably works too).



          There is also a more geometric approach to a definition like the one you've been given. If $Delta_n={(t_0,dots, t_n)mid 0leq t_ileq 1, sum_i t_i=1}$ is the standard $n$-simplex, a singular $n$-simplex on a topological space $X$ is a continuous map $f:Delta_nto X$. The graded set $S(X)$, where $S_n(X)$ is the set of singular $n$-simplices is called the total singular complex of $X$. If we define maps
          $$ (partial_i f)(t_0,dots, t_{n-1})=(t_0,dots, t_{i-1},0,t_i,dots, t_{n-1})$$
          $$ (s_i f)(t_0,dots, t_i+t_{i+1},dots, t_n)$$
          then $S(X)$ becomes a simplicial set. You can see that $s_i$ is precisely a component of your co-boundary map using additive notation. Simplicial objects play an important role in homology [see P. May, Simplicial Objects in Algebraic Topology]. In this case, $S_n(X)$ is not only a set but an abelian group, so we can define a boundary oparator as an alternating sum of $partial_i$, but one can also use $s_i$ to define a co-boundary operator.






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            $begingroup$

            This is more an extended comment that an answer, but I believe this might give you an idea of where the inspiration comes from.



            If $G$ is not a Lie group but a Lie algebra or a $k$-algebra and $P$ is $G$-bimodule, we can similarily define $C^p(G,P)=mathrm{Hom}(Gotimescdotsotimes G, P)$. When you define the cohomology like that it becomes a derived functor (see the examples). By this I mean that if you consider the functor $Z$ sending a $G$-bimodule $P$ to $Z(P)={xin Pmid gx=xg forall gin G}$, then the family $H^n(G,-)$ are the derived functors of $Z$. Unfortunately, I'm not sure how to translate this into the case when $G$ is not an algebra, maybe if you can obtain a $G$-bimodule-like (with only the group operation) on $mathbb{R}$ everything works (in the wikipedia link there is an example of group cohomology that probably works too).



            There is also a more geometric approach to a definition like the one you've been given. If $Delta_n={(t_0,dots, t_n)mid 0leq t_ileq 1, sum_i t_i=1}$ is the standard $n$-simplex, a singular $n$-simplex on a topological space $X$ is a continuous map $f:Delta_nto X$. The graded set $S(X)$, where $S_n(X)$ is the set of singular $n$-simplices is called the total singular complex of $X$. If we define maps
            $$ (partial_i f)(t_0,dots, t_{n-1})=(t_0,dots, t_{i-1},0,t_i,dots, t_{n-1})$$
            $$ (s_i f)(t_0,dots, t_i+t_{i+1},dots, t_n)$$
            then $S(X)$ becomes a simplicial set. You can see that $s_i$ is precisely a component of your co-boundary map using additive notation. Simplicial objects play an important role in homology [see P. May, Simplicial Objects in Algebraic Topology]. In this case, $S_n(X)$ is not only a set but an abelian group, so we can define a boundary oparator as an alternating sum of $partial_i$, but one can also use $s_i$ to define a co-boundary operator.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              This is more an extended comment that an answer, but I believe this might give you an idea of where the inspiration comes from.



              If $G$ is not a Lie group but a Lie algebra or a $k$-algebra and $P$ is $G$-bimodule, we can similarily define $C^p(G,P)=mathrm{Hom}(Gotimescdotsotimes G, P)$. When you define the cohomology like that it becomes a derived functor (see the examples). By this I mean that if you consider the functor $Z$ sending a $G$-bimodule $P$ to $Z(P)={xin Pmid gx=xg forall gin G}$, then the family $H^n(G,-)$ are the derived functors of $Z$. Unfortunately, I'm not sure how to translate this into the case when $G$ is not an algebra, maybe if you can obtain a $G$-bimodule-like (with only the group operation) on $mathbb{R}$ everything works (in the wikipedia link there is an example of group cohomology that probably works too).



              There is also a more geometric approach to a definition like the one you've been given. If $Delta_n={(t_0,dots, t_n)mid 0leq t_ileq 1, sum_i t_i=1}$ is the standard $n$-simplex, a singular $n$-simplex on a topological space $X$ is a continuous map $f:Delta_nto X$. The graded set $S(X)$, where $S_n(X)$ is the set of singular $n$-simplices is called the total singular complex of $X$. If we define maps
              $$ (partial_i f)(t_0,dots, t_{n-1})=(t_0,dots, t_{i-1},0,t_i,dots, t_{n-1})$$
              $$ (s_i f)(t_0,dots, t_i+t_{i+1},dots, t_n)$$
              then $S(X)$ becomes a simplicial set. You can see that $s_i$ is precisely a component of your co-boundary map using additive notation. Simplicial objects play an important role in homology [see P. May, Simplicial Objects in Algebraic Topology]. In this case, $S_n(X)$ is not only a set but an abelian group, so we can define a boundary oparator as an alternating sum of $partial_i$, but one can also use $s_i$ to define a co-boundary operator.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                This is more an extended comment that an answer, but I believe this might give you an idea of where the inspiration comes from.



                If $G$ is not a Lie group but a Lie algebra or a $k$-algebra and $P$ is $G$-bimodule, we can similarily define $C^p(G,P)=mathrm{Hom}(Gotimescdotsotimes G, P)$. When you define the cohomology like that it becomes a derived functor (see the examples). By this I mean that if you consider the functor $Z$ sending a $G$-bimodule $P$ to $Z(P)={xin Pmid gx=xg forall gin G}$, then the family $H^n(G,-)$ are the derived functors of $Z$. Unfortunately, I'm not sure how to translate this into the case when $G$ is not an algebra, maybe if you can obtain a $G$-bimodule-like (with only the group operation) on $mathbb{R}$ everything works (in the wikipedia link there is an example of group cohomology that probably works too).



                There is also a more geometric approach to a definition like the one you've been given. If $Delta_n={(t_0,dots, t_n)mid 0leq t_ileq 1, sum_i t_i=1}$ is the standard $n$-simplex, a singular $n$-simplex on a topological space $X$ is a continuous map $f:Delta_nto X$. The graded set $S(X)$, where $S_n(X)$ is the set of singular $n$-simplices is called the total singular complex of $X$. If we define maps
                $$ (partial_i f)(t_0,dots, t_{n-1})=(t_0,dots, t_{i-1},0,t_i,dots, t_{n-1})$$
                $$ (s_i f)(t_0,dots, t_i+t_{i+1},dots, t_n)$$
                then $S(X)$ becomes a simplicial set. You can see that $s_i$ is precisely a component of your co-boundary map using additive notation. Simplicial objects play an important role in homology [see P. May, Simplicial Objects in Algebraic Topology]. In this case, $S_n(X)$ is not only a set but an abelian group, so we can define a boundary oparator as an alternating sum of $partial_i$, but one can also use $s_i$ to define a co-boundary operator.






                share|cite|improve this answer











                $endgroup$



                This is more an extended comment that an answer, but I believe this might give you an idea of where the inspiration comes from.



                If $G$ is not a Lie group but a Lie algebra or a $k$-algebra and $P$ is $G$-bimodule, we can similarily define $C^p(G,P)=mathrm{Hom}(Gotimescdotsotimes G, P)$. When you define the cohomology like that it becomes a derived functor (see the examples). By this I mean that if you consider the functor $Z$ sending a $G$-bimodule $P$ to $Z(P)={xin Pmid gx=xg forall gin G}$, then the family $H^n(G,-)$ are the derived functors of $Z$. Unfortunately, I'm not sure how to translate this into the case when $G$ is not an algebra, maybe if you can obtain a $G$-bimodule-like (with only the group operation) on $mathbb{R}$ everything works (in the wikipedia link there is an example of group cohomology that probably works too).



                There is also a more geometric approach to a definition like the one you've been given. If $Delta_n={(t_0,dots, t_n)mid 0leq t_ileq 1, sum_i t_i=1}$ is the standard $n$-simplex, a singular $n$-simplex on a topological space $X$ is a continuous map $f:Delta_nto X$. The graded set $S(X)$, where $S_n(X)$ is the set of singular $n$-simplices is called the total singular complex of $X$. If we define maps
                $$ (partial_i f)(t_0,dots, t_{n-1})=(t_0,dots, t_{i-1},0,t_i,dots, t_{n-1})$$
                $$ (s_i f)(t_0,dots, t_i+t_{i+1},dots, t_n)$$
                then $S(X)$ becomes a simplicial set. You can see that $s_i$ is precisely a component of your co-boundary map using additive notation. Simplicial objects play an important role in homology [see P. May, Simplicial Objects in Algebraic Topology]. In this case, $S_n(X)$ is not only a set but an abelian group, so we can define a boundary oparator as an alternating sum of $partial_i$, but one can also use $s_i$ to define a co-boundary operator.







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Jan 1 at 15:08

























                answered Jan 1 at 15:00









                JaviJavi

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