unit element in cohomology and k-theory

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It is some well-known fact that there is the unit element in $K^-2(point)$ whereas there is no unit in $H^-2(point)$? Where it is come from?



For details I add the link http://pages.uoregon.edu/ddugger/kgeom.pdf , the last lines of the page 169. The author fixes this fact in the way that I also don't understand. Can anyone explain what is going on there?




differential-geometry algebraic-topology homology-cohomology characteristic-classes topological-k-theory




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    $H^-2$ is always zero. As for $K$-theory see ncatlab.org/nlab/show/Bott+element.
    – Qiaochu Yuan
    Aug 7 at 19:35










  • what means the notation $H^*[t,t^-1]$? what do square brackets are for?
    – Yelon
    Aug 7 at 20:15







  • 1




    The square brackets are used to denote a polynomial ring in the indiccated variables en.wikipedia.org/wiki/Polynomial_ring . A class in $H^*(X)[t,t^-1]$ is a finite polynomial $Sigma_i=-n,dots, n t^-ix_i$, where each $x_iin H^*(X)$ is a (possibly trivial) cohomology class.
    – Tyrone
    Aug 8 at 8:54











  • How from that conclude that for $*=0$ we have $H^ev(X)=oplus_iH^2i(X)$?
    – Yelon
    Aug 8 at 9:57














up vote
1
down vote

favorite












It is some well-known fact that there is the unit element in $K^-2(point)$ whereas there is no unit in $H^-2(point)$? Where it is come from?



For details I add the link http://pages.uoregon.edu/ddugger/kgeom.pdf , the last lines of the page 169. The author fixes this fact in the way that I also don't understand. Can anyone explain what is going on there?




differential-geometry algebraic-topology homology-cohomology characteristic-classes topological-k-theory




share|cite|improve this question















  • 1




    $H^-2$ is always zero. As for $K$-theory see ncatlab.org/nlab/show/Bott+element.
    – Qiaochu Yuan
    Aug 7 at 19:35










  • what means the notation $H^*[t,t^-1]$? what do square brackets are for?
    – Yelon
    Aug 7 at 20:15







  • 1




    The square brackets are used to denote a polynomial ring in the indiccated variables en.wikipedia.org/wiki/Polynomial_ring . A class in $H^*(X)[t,t^-1]$ is a finite polynomial $Sigma_i=-n,dots, n t^-ix_i$, where each $x_iin H^*(X)$ is a (possibly trivial) cohomology class.
    – Tyrone
    Aug 8 at 8:54











  • How from that conclude that for $*=0$ we have $H^ev(X)=oplus_iH^2i(X)$?
    – Yelon
    Aug 8 at 9:57












up vote
1
down vote

favorite









up vote
1
down vote

favorite











It is some well-known fact that there is the unit element in $K^-2(point)$ whereas there is no unit in $H^-2(point)$? Where it is come from?



For details I add the link http://pages.uoregon.edu/ddugger/kgeom.pdf , the last lines of the page 169. The author fixes this fact in the way that I also don't understand. Can anyone explain what is going on there?




differential-geometry algebraic-topology homology-cohomology characteristic-classes topological-k-theory




share|cite|improve this question











It is some well-known fact that there is the unit element in $K^-2(point)$ whereas there is no unit in $H^-2(point)$? Where it is come from?



For details I add the link http://pages.uoregon.edu/ddugger/kgeom.pdf , the last lines of the page 169. The author fixes this fact in the way that I also don't understand. Can anyone explain what is going on there?





differential-geometry algebraic-topology homology-cohomology characteristic-classes topological-k-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 7 at 19:29









Yelon

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  • 1




    $H^-2$ is always zero. As for $K$-theory see ncatlab.org/nlab/show/Bott+element.
    – Qiaochu Yuan
    Aug 7 at 19:35










  • what means the notation $H^*[t,t^-1]$? what do square brackets are for?
    – Yelon
    Aug 7 at 20:15







  • 1




    The square brackets are used to denote a polynomial ring in the indiccated variables en.wikipedia.org/wiki/Polynomial_ring . A class in $H^*(X)[t,t^-1]$ is a finite polynomial $Sigma_i=-n,dots, n t^-ix_i$, where each $x_iin H^*(X)$ is a (possibly trivial) cohomology class.
    – Tyrone
    Aug 8 at 8:54











  • How from that conclude that for $*=0$ we have $H^ev(X)=oplus_iH^2i(X)$?
    – Yelon
    Aug 8 at 9:57












  • 1




    $H^-2$ is always zero. As for $K$-theory see ncatlab.org/nlab/show/Bott+element.
    – Qiaochu Yuan
    Aug 7 at 19:35










  • what means the notation $H^*[t,t^-1]$? what do square brackets are for?
    – Yelon
    Aug 7 at 20:15







  • 1




    The square brackets are used to denote a polynomial ring in the indiccated variables en.wikipedia.org/wiki/Polynomial_ring . A class in $H^*(X)[t,t^-1]$ is a finite polynomial $Sigma_i=-n,dots, n t^-ix_i$, where each $x_iin H^*(X)$ is a (possibly trivial) cohomology class.
    – Tyrone
    Aug 8 at 8:54











  • How from that conclude that for $*=0$ we have $H^ev(X)=oplus_iH^2i(X)$?
    – Yelon
    Aug 8 at 9:57







1




1




$H^-2$ is always zero. As for $K$-theory see ncatlab.org/nlab/show/Bott+element.
– Qiaochu Yuan
Aug 7 at 19:35




$H^-2$ is always zero. As for $K$-theory see ncatlab.org/nlab/show/Bott+element.
– Qiaochu Yuan
Aug 7 at 19:35












what means the notation $H^*[t,t^-1]$? what do square brackets are for?
– Yelon
Aug 7 at 20:15





what means the notation $H^*[t,t^-1]$? what do square brackets are for?
– Yelon
Aug 7 at 20:15





1




1




The square brackets are used to denote a polynomial ring in the indiccated variables en.wikipedia.org/wiki/Polynomial_ring . A class in $H^*(X)[t,t^-1]$ is a finite polynomial $Sigma_i=-n,dots, n t^-ix_i$, where each $x_iin H^*(X)$ is a (possibly trivial) cohomology class.
– Tyrone
Aug 8 at 8:54





The square brackets are used to denote a polynomial ring in the indiccated variables en.wikipedia.org/wiki/Polynomial_ring . A class in $H^*(X)[t,t^-1]$ is a finite polynomial $Sigma_i=-n,dots, n t^-ix_i$, where each $x_iin H^*(X)$ is a (possibly trivial) cohomology class.
– Tyrone
Aug 8 at 8:54













How from that conclude that for $*=0$ we have $H^ev(X)=oplus_iH^2i(X)$?
– Yelon
Aug 8 at 9:57




How from that conclude that for $*=0$ we have $H^ev(X)=oplus_iH^2i(X)$?
– Yelon
Aug 8 at 9:57















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