A morphism of stacks \(f:\mathcal{D}\rightarrow \mathcal{C}\) is called a representable submersion if, for every morphism \(\underline{M}\rightarrow \mathcal{C}\), the fibred product \(\mathcal{D}\times_{\mathcal{C}}\underline{M}\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \(\mathcal{D}\times_{\mathcal{C}}\underline{M}\rightarrow M\) is a submersion. Following is a criterion for a map of stacks to be representable submersion. The result says it is enough to check for epimorphisms \(\underline{M}\rightarrow \mathcal{C}\). Precise statement is as follows. Let \(f:\mathcal{D}\rightarrow \mathcal{C}\) be a morphism of stacks. Suppose given a manifold \(U\) and a morphism of stacks \(\underline{U}\rightarrow \mathcal{C}\) which is an epimorphism. If the fibered product \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U\) is a submersion, then \(f\) is representable submersion. Let us see what this means in the set up of manifolds. Let \(f:\mathcal{D}\rightarrow \mathcal{C}\) be a morphism of stacks. Let \(f:M\rightarrow N\) be a morphism of manifolds (which gives a morphism of stacks \(\underline{M}\rightarrow \underline{N}\)). Suppose given a manifold \(U\) and a morphism of stacks \(\underline{U}\rightarrow \mathcal{C}\) which is an epimorphism. A representable surjective submersion is an epimorphism. So, we consider a surjective submersion \(g:U\rightarrow N\) (which gives an epimorphism \(\underline{U}\rightarrow \underline{N}\) being a representable surjective submersion). If the fibered product \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U\) is a submersion. As \(U\rightarrow N\) is submersion, it is anyways true that \(M\times_N U\) is a smooth manifold. What is extra that we have here is that \(M\times_NU\rightarrow U\) is a submersion. It is anyways true that \(M\times_NU\rightarrow M\) is a submersion being a pullback of submersion. But it is not true in general that \(M\times_NU\rightarrow U\) is a submersion. Here, we are given that \(M\times_NU\rightarrow U\) is a submersion. So, Let \(f:\mathcal{D}\rightarrow \mathcal{C}\) be a morphism of stacks. Suppose given a manifold \(U\) and a morphism of stacks \(\underline{U}\rightarrow \mathcal{C}\) which is an epimorphism. If the fibered product \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \(\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U\) is a submersion, then \(f\) is representable submersion. turns to Let \(f:M\rightarrow N\) be a morphism of manifolds be a morphism of manifolds. Suppose given a manifold \(U\) and a sujective submersion \(\underline{U}\rightarrow N\). If the fibered product \(M\times_{N}U\) is a manifold and that the map of manifolds \(M\times_{N}U\rightarrow U\) is a submersion, then \(f\) is submersion. This is more or less obvious. We have following commutative diagram As \(G:U\rightarrow N\) is a submersion (we have started with this) and \(p_2: M\times_N U\rightarrow U\) is a submersion (we are given this), the composition \(G\circ p_2=F\circ p_1\) is a submersion which then imply that \(F:M\rightarrow N\) is a submersion. Let \(m\in M\). As \(G:U\rightarrow N\) is surjective, so is \(p_1\) (pullback of surjective is surjective) i.e., there exists \((m,u)\in M\times_N U\) such that \(p_1(m,u)=m\). As \(F\circ p_1\) is submersion, \((F\circ p_1)_{*,(m,u)}(T_{m,u}(M\times_N U))=T_{F(m)}N\). Applying chain rule, we have \(F_{*,m}((p_1)_{*,(m,u)}(T_{m,u}(M\times_N U)))=T_{F(m)}N\), in particular, \(F_{*,m}(T_mM)=T_{F(m)}N\). Thus, \(F\) is submersion. So, we need both surjectivity and submersion of \(U\rightarrow N\). Now, let us look at more general case. Now, \(U\rightarrow N\) is not a surjective submersion but induces an epimorphism \(U\rightarrow N\). Suppose that the pullback \(M\times_N U\) is a manifold and that the map \(M\times_N U\rightarrow U\) is a submerson. Let \(W\rightarrow N\) be a map. We need to prove that \(M\times_N W\) is a manifold. We have following diagram As \(M\times_N U\rightarrow U\) is a submersion, the pullback \((M\times_N U)\times_U W_i=M\times_N W_i\) is a manifold. So, we have an open cover \(\{W_i\rightarrow W\}\) of \(W\) such that the pullbacks \(M\times_N W_i\) are manifolds. I think this should confirm that \(M\times_N W\) is a manifold and just because \(M\times_N W\rightarrow W_i\) are submersions, so is the map \(M\times_N W\rightarrow W\). Thus, \(f:M\rightarrow N\) is a representable submersion. The same idea works for an arbitrary map of stacks \(\mathcal{D}\rightarrow \mathcal{C}\). Let \(\underline{W}\rightarrow \mathcal{C}\) be a map of stacks. We have to prove that \(\mathcal{D}\times_{\mathcal{C}}\underline{W}\) is representable and that the map of manifolds \(\mathcal{D}\times_{\mathcal{W}}\underline{W}\rightarrow W\) is a submersion. As \(\underline{U}\rightarrow \mathcal{C}\) is epimorphism, for \(\underline{W}\rightarrow \mathcal{C}\) there exists an open cover \(\{W_i\rightarrow W\}\) with commutative diagram as shown below. We have following diagram As \(\mathcal{D}\times_{\mathcal{C}} \underline{U}\rightarrow U\) is a submersion, the pullback \((\mathcal{D}\times_{\mathcal{C}} \underline{U})\times_U W_i=\mathcal{D}\times_{\mathcal{C}} W_i\) is a manifold. So, we have an open cover \(\{W_i\rightarrow W\}\) of $ W$ such that the \((\mathcal{D}\times_{\mathcal{C}}W)\times_W W_i=\mathcal{D}\times_{\mathcal{C}} W_i\) ...

Let \(V\) be a vector space over \(\mathbb{R}\). Let \(\{e_1,\cdots,e_r\}\) be a basis of \(V\) over \(\mathbb{R}\) and \(\{e^1,\cdots,e^r\}\) be the dual basis of \(V\). We call \(e^i:V\rightarrow \mathbb{R}\) to be polynomials over \(V\) with values in \(\mathbb{R}\). A map \(p:V\rightarrow \mathbb{R}\) is said to be a polynomial map if \(p=\sum a_{t_1,\cdots,t_r}(e^1)^{t_1}\cdots(e^r)^{t_r}\). Let \(f:V\times V\times \cdots\times V\rightarrow \mathbb{R}\) be a symmetric multilinear mapping. We want to associate

Given a principal \(G\) bundle \(P\rightarrow M\) we associate what is called a Weil homomorphism \(I(G)\rightarrow H^*(M,\mathbb{R})\). Given \(f\in I^k(G)\) i.e., \(f:\underbrace{\mathfrak{g}\times\cdots\times\mathfrak{g}}_{k\text{ times}}\rightarrow \mathbb{R}\) we associate an element in \(H^{2k}(M,\mathbb{R})\) as follows. This is only an outline. It is useful if you can fill the gaps by your self. Fix a connection \(\Gamma\) on \(P(M,G)\) and let \(\Omega\) denote the curvature form associated to \(\Gamma\). The element \(f\in I^k(G)\) gives a \(2k\)-form \(f(\Omega):P\rightarrow \Lambda^{2k}T^*P\) on \(P\) as follows. \(f(\Omega)(v_1,\cdots,v_{2k})=\frac{1}{(2k)!}\sum_{\sigma\in S_{2k}} f(\Omega(v_{\sigma(1)}.v_{\sigma(2)}),\cdots\Omega(v_{\sigma(2k-1)},v_{\sigma(2k)}))\) ...

Let \(G\) be a Lie group and \(\mathfrak{g}\) be its Lie algebra. Let \(P\rightarrow M\) be a principal \(G\) bundle. A connection form on \(P\) is a \(\mathfrak{g}\) valued \(1\)-form on \(P\) satisfying some properties. Suppose \(G=Gl(n,\mathbb{R})\) then \(\mathfrak{g}=M(n,\mathbb{R})\). A connection is given by \(\omega:P\rightarrow \Lambda^1_{\mathfrak{g}}T^*P\). Given \(p\in P\) we have \(\omega(p):T_pP \rightarrow \mathfrak{g}\). Given \(v\in T_pP\), \(\omega(p)(v)\) is a matrix \((a_{ij})\in M(n,\mathbb{R})\) i.e., given \(v\in T_pP\) we have \(n^2\) real numbers \(a_{ij}\in \mathbb{R}\) associated to it. Varying \(v\) over \(T_pP\) gives \(n^2\) maps \(a_{ij}:T_pP\rightarrow \mathbb{R}\). So, given \(p\in P\), we have \(n^2\) maps \(\omega_{ij}(p):T_pP\rightarrow \mathbb{R}\) where \(\omega_{ij}(p)(v)\) is the \(ij\) th component of \(\omega(p)(v)\). Fix \(i,j\) then, \(\omega_{ij}:P\rightarrow \Lambda^1 T^*P\) given by \(p\mapsto \omega_{ij}(p)\) is a real valued \(1\)-form on \(P\). Thus, we denote \(\omega\) by \((\omega_{ij})\) where \(\omega_{ij}\) are real valued \(1\)-forms on \(P\). This is what it means to see connection as a matrix of \(1\)-forms. The same can be done for Curvature form also. Curvature form \(\Omega:P\rightarrow \Lambda^2_{\mathfrak{g}}TP\) associates for each \(p\in P\) a map \(\Omega(p):T_pP\times T_pP\rightarrow \mathfrak{g}\). Same explanation as above gives \(n^2\) real valued \(2\)-forms \(\Omega_{ij}:P\rightarrow \Lambda^2 TP\). We denote Curvature form \(\Omega\) by \((\Omega_{ij})\). This is what it means to see curvature as a matrix of \(2\)-forms.

Here, I will add links for web pages where I have written about concepts from Kobayashi and Nomizu's book Foundations of Differential geometry (Volume \(1\) and Volume \(2\)). Derivative of Left invariant differential form Maurer-Cartan form on a Lie group Transition maps for principal bundle are smooth Trivializations and sections in Principal bundle Construction of Weil homomorphism Construction of associated bundle Equivariant maps are Isomorphisms Invariant polynomials

In this, we see that for a left-invarinat differential form \(\omega\) on $G$, \(d\omega(X,Y)=-\frac{1}{2}\omega([X,Y])\) for vector fields \(X, Y\) in \(G\). Let \(\omega:G\rightarrow \Lambda^1 T^*G\) be a Left-invariant differential form on \(G\) i.e., \((L_g)^*\omega=\omega\) for all \(g\in G\) i.e., \(\omega(g)(v)=\omega(e)((L_{g^{-1}})_{*,g}(v))\) for all \(g\in G\) and \(v\in T_gG\). Given \(A\in \mathfrak{g}\) we have vector field \(A^*:G\rightarrow TG\) defined as \(A^*(g)=(L_g)_{*,e}(A)\). Then, \(\omega(A^*):G\rightarrow \mathbb{R}\) is constant function, \(\omega(A^*)(g)=\omega(e)(A)\) for all \(g\in G\). As \(\omega(A^*):G\rightarrow \mathbb{R}\) is constant map, \(X( \omega(A^*))=0\) for any vector field \(X:G\rightarrow TG\) on \(G\). In particular, \(B^*(X(\omega(A^*)))=0\) for \(B\in \mathfrak{g}\). Interchanging \(A\) and \(B\) we have \(A^*(\omega(B^*))=0\) As \((d\omega)(A^*,B^*)=\frac{1}{2}\left[ A^*(\omega(B^*))-B^*(\omega(A^*))-\omega([A^*,B^*])\right]\) ...

Let \(G\) be a Lie group and \(\mathfrak{g}\) be its Lie algebra. We want to associate a \(\mathfrak{g}\) valued \(1\) form on \(G\). We define \(\theta:G\rightarrow \Lambda^1_{\mathfrak{g}}T^*G\) as follows. For \(g\in G\), we need \(\theta(g):T_gG\rightarrow \mathfrak{g}=T_eG\). For manifolds \(M,N\), one natural way to get a map between tangent spaces \(T_mM\) and \(T_nN\) is to think of a smooth map \(f:M\rightarrow N\) such that \(f(m)=n\) and take its differential at \(m\). We get \(f_{*,m}:T_mM\rightarrow T_nN\). To get \(\theta(g):T_gG\rightarrow \mathfrak{g}=T_eG\), we look for a map \(G\rightarrow G\) that takes \(g\) to \(e\). One such map is multiplication by \(g^{-1}\). Consider \(\delta_{g^{-1}}:G\rightarrow G\) given by \(h\mapsto g^{-1}h\). This map takes \(g\) to \(e\) and \(\delta_{*,g^{-1}}:T_gG\rightarrow T_eG=\mathfrak{g}\). This gives a \(\mathfrak{g}\) valued \(1\)-form on \(G\) which we call to be the Maurer-Cartan form on \(G\) denoted by \(\theta\) defined as \(\theta(g)=(\delta_{g^{-1}})_{*,g}:T_gG\rightarrow \mathfrak{g}\). Kobayashi and Nomizu defines Maurer-Cartan form on \(G\) to be "the left-invariant \(\mathfrak{g}\) valued \(1\)-form on \(G\) uniquely determined by the condition that \(\theta(A)=A\) for all \(A\in \mathfrak{g}\). More precisely, this means \(\theta (e) :T_eG\rightarrow T_eG\) is such that \(\theta(e)(A)=A\) for all \(A\in \mathfrak{g}\). The condition that \(\theta\) is left-invariant means that \(\theta(e):T_eG\rightarrow \mathfrak{g}\) determines what \(\theta(g):T_gG\rightarrow \mathfrak{g}\). So, the condition \(\theta(e)(A)=A\) for all \(A\in \mathfrak{g}\) along with the condition left-invariant gives unique \(1\)-form \(\theta\) which is called as the Maurer-Cartan form. Once we unravel how \(\theta(e):T_eG\rightarrow \mathfrak{g}\) determines \(\theta(g):T_gG\rightarrow \mathfrak{g}\) and that \(\theta(e)(A)=A\) for all \(A\in \mathfrak{g}\) we see that \(\theta(g)(v)=(\delta_{g^{-1}})_{*,g}(v)\) for all \(v\in T_gG\) which is precisely what I have written in the first half of this post. Differentiating this Maurer-Cartan form \(\theta:G\rightarrow \Lambda^1_{\mathfrak{g}}T^*G\) we see that \(d\theta(X,Y)=-\frac{1}{2}\theta([X,Y])\) ...

Definition : Let \(\mathcal{C}\) be a category and \(f:A\rightarrow C\) and \(g:B\rightarrow C\) be morphisms. We define pull back of \(f,g\) to be a triple \((P,f',g')\) where \(P\) is an object of \(\mathcal{C}\) and \(f':P\rightarrow B, g':P\rightarrow A\) with \(g\circ f'=f\circ g'\) such that given any object \(P'\) of \(\mathcal{C}\) and morphisms \(g'':P'\rightarrow A\) and \(f'':P'\rightarrow B\) with \(g\circ f''=f\circ g''\), there exists a unique morphism \(\eta:P'\rightarrow P\) such that \(g''=g'\circ \eta\) and \(f''=f'\circ \eta\) as in the following commutative diagram. Definition : Let \(\mathcal{C}\) be a category and \(f:A\rightarrow C\) and \(g:B\rightarrow C\) be morphisms. We define pull back of \(f,g\) to be a triple \((P,f',g')\) where \(P\) is an object of \(\mathcal{C}\) and \(f':P\rightarrow B, g':P\rightarrow A\) with \(g\circ f'=f\circ g'\) such that given any object \(P'\) of \(\mathcal{C}\) and morphisms \(g'':P'\rightarrow A\) and \(f'':P'\rightarrow B\) with \(g\circ f''=f\circ g''\), there exists a unique morphism \(\eta:P'\rightarrow P\) such that \(g''=g'\circ \eta\) and \(f''=f'\circ \eta\) as in the following commutative diagram.