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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Complex geometry\\[15mm] \small lecture 8: Levi-Civita connection} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] October 17, 2020 } \end{center} \newpage {\bf \blue Connections} \definition Recall that {\bf \blue a connection} on a bundle $B$ is an operator $\nabla:\; B \arrow B \otimes \Lambda^1 M$ satisfying $\nabla(fb) = b \otimes df + f \nabla(b)$, where $f \arrow df$ is de Rham differential. When $X$ is a vector field, we denote by $\nabla_X(b)\in B$ the term $\langle \nabla(b), X\rangle$. \remark {\bf \purple A connection $\nabla$ on $B$ gives a connection $B^* \stackrel {\nabla^*} \arrow \Lambda^1 M \otimes B^*$ on the dual bundle,} by the formula \[ d(\langle b, \beta\rangle) = \langle \nabla b, \beta\rangle+ \langle b, \nabla^*\beta\rangle \] These connections are usually denoted {\bf \red by the same letter $\nabla$.} \remark For any tensor bundle ${\cal B}_1:= B^*\otimes B^* \otimes ... \otimes B^* \otimes B\otimes B \otimes ... \otimes B$ {\bf \purple a connection on $B$ defines a connection on ${\cal B}_1$} using the Leibniz formula: \[ \nabla(b_1 \otimes b_2) = \nabla(b_1) \otimes b_2 + b_1 \otimes \nabla(b_2). \] \newpage {\bf \blue Parallel transport along the connection} \theorem Let $B$ be a vector bundle with connection over $\R$. Then for each $x\in \R$ and each vector $b_x \in B\restrict x$ {\bf \red there exists a unique section $b\in B$ such that $\nabla b=0$, $b\restrict x= b_x$.} \proof This is existence and uniqueness of solutions of an ODE $\frac {db}{dt} + A(b)=0$. \endproof \definition Let $\gamma:\; [0, 1] \arrow M$ be a smooth path in $M$ connecting $x$ and $y$, and $(B, \nabla)$ a vector bundle with connection. Restricting $(B, \nabla)$ to $\gamma([0,1])$, we obtain a bundle with connection on an interval. Solve an equation $\nabla(b)=0$ for $b\in B\restrict{\gamma([0,1])}$ and initial condition $b\restrict x= b_x$. This process is called {\bf\blue parallel transport} along the path via the connection. The vector $b_y:= b\restrict y$ is called {\bf\blue vector obtained by parallel transport of $b_x$ along $\gamma$}. {\bf\blue Holonomy group} of $\gamma$ is the group of endomorphisms of the fiber $B_x$ obtained from parallel transports along all paths starting and ending in $x\in M$ \newpage {\bf \blue Parallel tensors} \definition Let $B$ be a vector bundle, and $\Psi\in B^{\otimes i} \otimes (B^*)^{\otimes j}$ a tensor on $B$. We say that {\bf \blue connection $\nabla$ preserves $\Psi$} if $\nabla(\Psi)=0$. In this case we also say that $\Psi$ is {\bf \blue parallel} with respect to the connection. \remark $\nabla(\Psi)=0$ is equivalent to $\Psi$ being a solution of $\nabla(\Psi)=0$ on each path $\gamma$. This means that {\bf \purple parallel transport preserves $\Psi$.} We obtained \corollary {\bf \red A tensor is parallel if and only if it is holonomy invariant.} \example {\bf \blue Orthogonal connection}: given a positive definite form $h\in \Sym^2 B^*$ on $B$, a connection $\nabla$ such that $\nabla(h)=0$ is called {\bf \blue orthogonal}. \example Suppose that $(B,I)$ is a complex vector bundle equipped with a Hermitian metric $h$. A connection $\nabla$ such that $\nabla(I)=\nabla(h)=0$ is called {\bf \blue unitary}, or {\bf \blue Hermitian}. \newpage {\bf \blue Torsors and affine spaces} \definition {\bf \blue A torsor} over a group $G$ is a space $X$ equipped with a free and transitive action of $G$, $g,x \arrow \rho(g,x).$ \definition {\bf \blue Morphism} of torsors $(X,G,\rho) \stackrel \Psi \arrow (X',G',\rho')$ is a pair $\Psi_X:\; X\arrow X', \Psi_G:\; G \arrow G'$, where $\Psi_G$ is a group homomorphism satisfying $\Psi_X(\rho(g,x)) = \rho'(\Psi_G(g),\Psi_X(x))$ (that is, compatible with the map $\Psi_X$). \remark {\bf \red This defines the category of torsors.} \definition {\bf \blue Affine space} is a torsor over a vector space $V$, which is called {\bf \blue linearization}. The action of $V$ on $A$ is denoted $a,v \arrow a +v$. \example Given two connections $\nabla$ and $\nabla_1$ on $B$, the difference $\nabla-\nabla_1$ is an $\End(B)$-valued 1-form. Converse is also true: for any $\End(B)$-valued 1-form $A\in \Lambda^1M \otimes \End(B)$, the operator $\nabla+A$ is a connection. In other words, {\bf \red the space of connections is an affine space over $\Lambda^1M \otimes \End(B)$}. \newpage {\bf \blue Affine space of orthogonal connections} \claim Let $B$ be a bundle with a scalar product. Then {\bf \red the space of orthogonal connections on $B$ an affine space over $\Lambda^1 M \otimes \goth{so}(B)$.} \proof Let $s\in B^*\otimes B^*$ be a 2-form on $B$. The action of $A:= \nabla-\nabla_1$ on $B^*\otimes B^*$ is given by $A(s) (x,y)= - s(A(x), y) - s(x, A(y))$. Therefore, a difference $A$ of orthogonal connections satisfies $h(A(x), y) = - h(x, A(y))$ for all $x, y\in B$. This is the same as $A\in \Lambda^1 M \otimes \goth{so}(B)$. \endproof Similarly one proves \claim Let $B$ be a bundle with a Hermitian structure and a tensor $\Phi$, and ${\goth g}\subset \End(B)$ the Lie algebra of endomorphisms preserving $\Phi$. Then {\bf \red the space of connections on $B$ preserving $\Phi$ is an affine space over $\Lambda^1 M \otimes \g$.} \newpage {\bf \blue REMINDER: de Rham algebra} \definition Let $\Lambda^* M$ denote the vector bundle with the fiber $\Lambda^*T^*_xM$ at $x\in M$ ($\Lambda^*T^*M$ is the Grassman algebra of the cotangent space $T^*_x M$). The sections of $\Lambda^i M$ are called {\bf \blue differential $i$-forms}. The algebraic operation ``wedge product'' defined on differential forms is $C^\infty M$-linear; the space $\Lambda^* M$ of all differential forms is called {\bf \blue the de Rham algebra}. \remark $\Lambda^0 M = C^\infty M$. \theorem {\bf \red There exists a unique operator $C^\infty M\stackrel d \arrow \Lambda^1 M\stackrel d \arrow \Lambda^2 M \stackrel d \arrow \Lambda^3 M \stackrel d \arrow ...$ satisfying the following properties} 1. On functions, $d$ is equal to the differential.\\ 2. $d^2=0$ \\ 3. $d(\eta \wedge \xi) = d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge d(\xi)$, where $\tilde \eta=0$ where $\eta\in \lambda^{2i}M$ is {\bf \blue an even form,} and $\eta\in \lambda^{2i+1}M$ is {\bf \blue odd.} \definition The operator $d$ is called {\bf \blue de Rham differential}. \newpage {\bf \blue Cartan formula} \claim For any $\eta \in \Lambda^1 M$, and $X,Y\in TM$ one has {\bf \red\[ d\eta(X,Y) = \eta([X,Y])- \Lie_X(\eta(Y))+ \Lie_Y(\eta(X)). \]} \phantom{x} \hspace{-14mm} \proof Two sides of this equation define two operators $d, d_1\Lambda^1 M \arrow \Lambda^2 M$. Both operators satisfy the Leibniz rule $d(f\eta)= df\wedge d\eta + f d\eta$. When $\eta=df$ is exact, one has \begin{multline*} \eta([X,Y])- \Lie_X(\eta(Y))+ \Lie_Y(\eta(X)) = \\ = \Lie_{[X,Y]}(f)-\Lie_X\Lie_Y(f)+\Lie_Y\Lie_X(f)=0 \end{multline*} hence $d_1(\alpha)=0$ on all closed forms. A map $\delta:\; \Lambda^1(M)\arrow \Lambda^2(M)$ which vanishes on closed forms and satisfies the Leibniz rule is de Rham differential, which can be seen from the axiomatic definition of $d$. \endproof \newpage {\bf \blue Torsion} \definition Let $\nabla$ be a connection on $\Lambda^1 M$, \[ \Lambda^1 \stackrel \nabla \arrow \Lambda^1 M \otimes \Lambda^1M.\] {\bf \blue Torsion of $\nabla$} $T_\nabla:\; \Lambda^1 M \arrow \Lambda^2 M$ is a map $\nabla \circ \Alt - d$, where $\Alt:\; \Lambda^1 M \otimes \Lambda^1M\arrow \Lambda^2 M$ is exterior multiplication. \remark \begin{align*} T_\nabla(f\eta) = & \Alt(f\nabla\eta + df\otimes \eta) - d(f\eta)\\ = &f\bigg [\Alt(\nabla\eta) - d\eta\bigg] + df\wedge \eta - df\wedge \eta= f T_\nabla(\eta). \end{align*} {\bf \purple Therefore $T_\nabla$ is linear}. \definition Let $(M,g)$ be a Riemannian manifold. A connection $\nabla$ on $TM$ is called {\bf \blue orthogonal} if $\nabla(g)=0$, and {\bf \blue Levi-Civita connection} if it is orthogonal and has zero torsion. \theorem {\bf \blue (``the fundamental theorem of Riemannian geometry'')} {\bf \red Every Riemannian manifold admits a Levi-Civita connection, and it is unique.} {\bf \green Will be proven later today}. \newpage {\bf \blue Gregorio Ricci-Curbastro, Tullio Levi-Civita} {\setlength\tabcolsep{20mm} \begin{tabular}{cc} \epsfig{file=Ricci-Curbastro.jpeg,width=0.32\linewidth} & \epsfig{file=Levi-Civita_3.jpeg,width=0.35\linewidth}\\ Gregorio Ricci-Curbastro, & Tullio Levi-Civita, \\ 1853-1925 & 1873-1941 \end{tabular} } {\green \em ...With his former student Tullio Levi-Civita, he wrote his most famous single publication, a pioneering work on the calculus of tensors, signing it as Gregorio Ricci. This appears to be the only time that Ricci-Curbastro used the shortened form of his name in a publication, and continues to cause confusion.} \newpage {\bf \blue Torsion and commutator of vector fields} \remark Cartan formula gives \begin{align*} T_\nabla(\eta)(X,Y) = &\nabla_X(\eta)(Y) - \nabla_Y(\eta)(X)- d\eta(X,Y) \\ =& \nabla_X(\eta)(Y) - \nabla_Y(\eta)(X) -\eta([X,Y])- \Lie_X(\eta(Y))+ \Lie_Y(\eta(X)). \end{align*} On the other hand, $\nabla_X(\eta)(Y)= \Lie_X(\eta(Y)) - \eta(\nabla_X(Y))$. Comparing the equations, we obtain \[ T_\nabla(\eta)(X,Y)=\eta\bigg(\nabla_X(Y)- \nabla_Y(X) - [X,Y]\bigg). \] {\bf \blue Torsion is often defined as a map $\Lambda^2 TM \arrow TM$ using the formula $\nabla_X(Y)- \nabla_Y(X) - [X,Y]$.} We have just proved \claim {\bf \red The torsion tensor $\nabla_X(Y)- \nabla_Y(X) - [X,Y]$ is dual to the torsion $\nabla \circ \Alt - d:\; \Lambda^1 M \arrow \Lambda^2 M$ defined above}. \endproof \newpage {\bf \blue Linearization of the torsion} \remark Consider the space ${\cal A}(\Lambda^1 M)$ of connections on $\Lambda^1 M$. The torsion defines an affine map \[ {\cal A}(\Lambda^1 M) \arrow \Hom(\Lambda^1M, \Lambda^2 M)= TM \otimes\Lambda^2 M . \] because $T(\nabla + \alpha) = T(\nabla) + \Alt_{12}(\alpha)$, where $\Alt_{12}:\; \Lambda^1M \otimes \End(\Lambda^1M) \arrow \Lambda^2M \otimes TM$ is antisymmetrization in the first two indices. \definition {\bf \blue Liearized torsion} is a map \[ T_{lin}:\; \Lambda^1(M) \otimes \Lambda^1(M) \otimes TM \arrow \Lambda^2 M \otimes TM \] obtained as a linearization of the torsion map. {\bf \purple It is equal to $\Alt_{12}$.} \newpage {\bf \blue Existence of orthogonal connections} \claim Let $B$ be a vector bundle equipped with a scalar product. {\bf \red Then $B$ admits an orthogonal connection}. \proof Chose a covering $\{U_i\}$, such that $B$ is trivial on each $U_i$ and admits an orthonormal basis in each $U_i$. On each $U_i$ we chose a connection $\nabla_i$ preserving this basis. Let $\psi_i$ be a partition of unit subjugated to $\{U_i\}$. Then {\bf \purple the formula $\nabla(b):= \sum \nabla_i(\psi_i b)$ defines an orthogonal connection}. \theorem {\bf \blue (``the fundamental theorem of Riemannian geometry'')} {\bf \red Every Riemannian manifold admits a Levi-Civita connection, and it is unique.} \proof See the next slide. \newpage {\bf \blue Levi-Civita connection (existence and uniqueness)} \pstep Chose an orthogonal connection $\nabla_0$ on $\Lambda^1 M$. The space ${\cal A}$ of orthogonal connections is affine and {\bf \purple its linearization is $\Lambda^1 M \otimes {\goth{so}}(TM)$.} We shall identify $\goth{so}(TM)$ and $\Lambda^2 M$. Then {\bf \purple ${\cal A}$ is an affine space over $\Lambda^1 M \otimes \Lambda^2 M$}. {\bf \green Step 2:} Then the linearized torsion map is \[ T_{lin} :\; \Lambda^1 M \otimes {\goth {so}}(TM)= \Lambda^1(M) \otimes \Lambda^2 M \stackrel{\Alt_{12}} \arrow \Lambda^2 M \otimes \Lambda^1M = \Lambda^2 M \otimes T M. \] {\bf \purple It is an isomorphism}. Indeed, on the right and on the left there are bundles of the same rank, hence it would suffice to show that $T_{lin}=\Alt_{12}$ is injective. However, if $\eta \in \ker T_{lin}$, it is a form which is symmetric on first two arguments and antisymmetric on the second two, giving $\eta(x,y,z) = \eta(y,x,z) = - \eta (y,z, x).$ This gives $\sigma(\eta) =-\eta$, where $\sigma$ is a cyclic permutation of the arguments. Since $\sigma^3=1$, this implies $\eta=0$. {\bf \green Step 3:} We have shown that {\bf \purple an orthogonal connection is uniquely determined by its torsion}. Indeed, torsion map is an isomorphism of affine spaces. {\bf \green Step 4:} Let $\nabla:= \nabla_0 -T_{lin}^{-1}(T_{\nabla_0})$. Then $T_\nabla= T_{\nabla_0}-T_{lin}(T_{lin}^{-1}(T_{\nabla_0}))=0$, hence {\bf \red $\nabla$ is torsion-free.} \endproof \end{document}