Кіріспе
Дұрыс гомотопиялық жағдай
Exact homotopy case
Математикада, U(n) бірлік тобының жіктеу кеңістігі BU(n) кеңістігімен және әмбебап бунделі EU(n) бірге келеді, осылайша, паракомпакт кеңістіктегі кез келген эрмиттік бундель X, X → BU(n) картасы арқылы EU(n) бунделінің кері тартылуы болады, және гомотопияға дейін бірегей. Бұл кеңістік, өзінің әмбебап фибрациясымен бірге, шексіз өлшемді кешенді Гильберт кеңістігіндегі n жазықтықтың Грассманнианы ретінде немесе n жазықтықтың Грассманниандарының индукцияланған топологиясымен тікелей лиміті ретінде құрылуы мүмкін. Екі құрылым да осы жерде егжей-тегжейлі сипатталған.
In mathematics, the classifying space for the unitary group U(n) is a space BU(n) together with a universal bundle EU(n) such that any hermitian bundle on a paracompact space X is the pull back of EU(n) by a map X → BU(n) unique up to homotopy. This space with its universal fibration may be constructed as either
the Grassmannian of n planes in an infinite dimensional complex Hilbert space; or,
the direct limit, with the induced topology, of Grassmannians of n planes. Both constructions are detailed here.
the Grassmannian of n planes in an infinite dimensional complex Hilbert space; or,
the direct limit, with the induced topology, of Grassmannians of n planes. Both constructions are detailed here.
Индуктивті шек ретінде құрылыс
Fn(Ck) – Ck-дегі n векторлардың ортонормалдық отбасыларының кеңістігі, ал Gn(Ck) – Ck-дің n өлшемді кіші кеңістіктерінің Грассман кеңістігі. Универсалды шоғырдың толық кеңістігі k → ∞ болғанда Fn(Ck) тікелей шегі ретінде алынуы мүмкін, ал базалық кеңістік k → ∞ болғанда Gn(Ck) тікелей шегі ретінде алынуы мүмкін.
Let Fn(Ck) be the space of orthonormal families of n vectors in Ck and let Gn(Ck) be the Grassmannian of n dimensional subvector spaces of Ck. The total space of the universal bundle can be taken to be the direct limit of the Fn(Ck) as k → ∞, while the base space is the direct limit of the Gn(Ck) as k → ∞.
Құрылыстың жарамдылығы
Бұл бөлімде біз EU(n) топологиясын анықтаймыз және EU(n) шын мәнінде келісімге ие екенін дәлелдейміз. U(n) тобы Fn(Ck) бойынша еркін әрекет етеді, ал коэффициент Грасмандық Gn(Ck) болып табылады. Келесі карта
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Fn−1(Ck−1) талшықтың талшықтық шоғыры болып табылады. Осылайша, k жеткілікті үлкен болғандықтан, және фибрацияның ұзын толық тізбегінен, біз кез келген кезде
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
, дәл , процесті қайталай аламыз және аламыз.
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Бұл соңғы топ k > n + p үшін тривиальды. Енді
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Fn(Ck) барлық тікелей шегі болсын (индуцирленген топологиямен).
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Gn(Ck) барлық тікелей шегі болсын (индуцирленген топологиямен).
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Лемма: топ барлық p ≥ 1 үшін тривиальды.
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Дәлел: γ : Sp → EU(n болсын, Sp компакт болғандықтан, γ(Sp) Fn(Ck) ішіне кіретіндей k бар. k-ні жеткілікті үлкен деп таңдасақ, γ негізгі нүктеге қатысты тұрақты картаға гомотопты екенін көреміз. Сонымен қатар, U(n) тобы EU(n) бойынша еркін әрекет етеді. Fn(Ck) және Gn(Ck) кеңістіктері CW кешендері болып табылады. Бұл кеңістіктерді CW кешендеріне жіктеуге болады, яғни Fn(Ck) үшін, сәйкесінше Gn(Ck) үшін, Fn(Ck+1) үшін біреуін шектеу арқылы, сәйкесінше Gn(Ck+1). Осылайша EU(n) (сондай-ақ Gn(C∞)) – CW кешені. Уайтхед теоремасы және жоғарыдағы лемма бойынша, EU(n) келісімге ие.
In this section, we will define the topology on EU(n) and prove that EU(n) is indeed contractible. The group U(n) acts freely on Fn(Ck) and the quotient is the Grassmannian Gn(Ck). The map
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
is a fibre bundle of fibre Fn−1(Ck−1). Thus because is trivial and because of the long exact sequence of the fibration, we have
whenever By taking k big enough, precisely for , we can repeat the process and get
This last group is trivial for k > n + p. Let
be the direct limit of all the Fn(Ck) (with the induced topology). Let
be the direct limit of all the Gn(Ck) (with the induced topology). Lemma: The group is trivial for all p ≥ 1. Proof: Let γ : Sp → EU(n), since Sp is compact, there exists k such that γ(Sp) is included in Fn(Ck). By taking k big enough, we see that γ is homotopic, with respect to the base point, to the constant map. In addition, U(n) acts freely on EU(n). The spaces Fn(Ck) and Gn(Ck) are CW complexes. One can find a decomposition of these spaces into CW complexes such that the decomposition of Fn(Ck), resp. Gn(Ck), is induced by restriction of the one for Fn(Ck+1), resp. Gn(Ck+1). Thus EU(n) (and also Gn(C∞)) is a CW complex. By Whitehead Theorem and the above Lemma, EU(n) is contractible.
Сансыз жіктеу кеңістігі
Каноникалық енгізулер сәйкес классификациялық кеңістіктерінде каноникалық енгізулерді тудырады. Олардың тиісті колимиттері былай белгіленеді: шындығында , бұл - кеңістігінің классификациялық кеңістігі.
The canonical inclusions induce canonical inclusions on their respective classifying spaces. Their respective colimits are denoted as:
is indeed the classifying space of .
is indeed the classifying space of .