It is true that if connection $\omega$ in a vector bundle is $\mathfrak{g}$-valued ($\mathfrak{g}$ being Lie Algebra of the structure Lie Group $G$) in a patch $U$, then it will be $\mathfrak{g}$-valued in all other patches due to the transformation law:

\begin{align} \omega_V = c^{-1}_{UV} \omega_U c_{UV} + c^{-1}_{UV} d c_{UV}. \end{align}

However, I'm trying to understand if it must take values in Lie algebra in order to be a connection?

  • $\begingroup$ How do you define a connection? The references I have seen just define the connection to be $\frak g$-valued $\endgroup$
    – Aweygan
    Commented Jul 25, 2019 at 20:42
  • $\begingroup$ I'm studying using T. Frankel's textbook. He defined the connection using its transformational properties there, but suddenly in later chapters often assumes that the connection is $\mathfrak{g}$-valued which I find very confusing. $\endgroup$
    – Darkseid
    Commented Jul 25, 2019 at 21:33
  • $\begingroup$ Clarification: the connection is defined as a $K \times K$ matrix of 1-forms that transforms according to the rule above, where $K$ comes from the rank of the vector bundle. $\endgroup$
    – Darkseid
    Commented Jul 25, 2019 at 22:01

3 Answers 3


Yes, your connection needs to be valued in the Lie algebra $\mathfrak g$ of a Lie group $G$ which is the structure group of your bundle ($ G = GL(n)$ for a rank-$n$ vector bundle, $G = O(n)$ for a rank-$n$ real vector bundle with metric, etc). Here is the intuition behind that:

You choose some patch $U$ and use a local trivialization to write your vector bundle as $V \mid_U \cong U \times \mathbb R^n$. Via this local trivialization, you have identified all the fibers $V_x, x \in U$ with $\mathbb R^n$. So now you can write sections of $V$ over $U$ as $\mathbb R^n$-valued functions and multivariable calculus tells you how to take partial derivatives along given directions.

But you should not just naively start taking partial derivatives of the individual components of a section. There are many possible trivializations of $V$ over $U$, and if your friend chooses a different trivialization and you both take partial derivatives in the naive way, you will disagree with each other. This is why we need covariant derivatives.

So the connection enters: We recognize that in addition to the naive recipe for taking partial derivatives, we need to add on a "rotation" of the section as we take derivatives along a given direction. So say that at the point $x\in U$ we want to take a covariant derivative along the $X \in T_xM$ direction (and say $s$ is our section over $U$): We take "naive" multivariable calculus derivatives of the components of $s$ in the $X$ direction and then we ask our Lie algebra-valued connection 1-form $\omega$ how we should "rotate" $s$ in order to make the derivative covariant.

Now $\omega$ eats the tangent direction $X$ and gives us a Lie algebra element, $\omega(X) \in \mathfrak g$. Elements of $\mathfrak g$ are (intuitively speaking) infinitesimal "rotations," where "rotation" means "element of $G$." So we act on $s$ by the infinitesimal rotation $\omega(X)$ and we add the result to the "naive derivative" of $s$ along the $X$-direction.

(Above "let $\omega(X)$ act on $s$" meant implicitly that we have fixed some representation of $G$ on the fibers of $V$ so that an expression like $\omega(X) s$ makes sense.)

So the concise answer: $\omega$ has to be Lie-algebra valued because its job is to specify infinitesimal rotations that we add to the usual partial derivatives in order to make the derivative covariant.

I haven't attempted to convince you that this procedure will resolve the original difficulty of consistently defining derivatives of sections despite multiple ways of trivializing $V$ over $U$, but that's the essence of the transformation law you wrote down. If you'd like I could edit this answer with more detail along those lines.

I haven't read your reference by Frankel but I recommend that you additionally look at DuPont's Fibre Bundles and Chern-Weil Theory.

  • $\begingroup$ Would it be fair to say that the crux of your answer is that the connection must be trivially zero in at least one coordinate patch, and thus due to the transformation law it can take values only in $\mathfrak{g}$ in other patches? $\endgroup$
    – Darkseid
    Commented Jul 26, 2019 at 10:24
  • $\begingroup$ Well I can't completely agree with that--I wouldn't say it "must be zero in at least one coordinate patch." It's possible for it to be nonzero in all coord. patches. But yes, the fact that you will be discussing multiple coord. patches and also want that transformation law will force it to be Lie alg-valued. $\endgroup$
    – Todd N
    Commented Jul 27, 2019 at 12:51
  • $\begingroup$ I would say the crux of my answer is that in the world of vector bundles, the right derivative to use is no longer a simple difference quotient. It's a difference quotient along with a term that accounts for the fact that the fibers can be "twisted" relative to each other. That term is an infinitesimal rotation and these are encoded by Lie algebra elements. $\endgroup$
    – Todd N
    Commented Jul 27, 2019 at 12:53
  • $\begingroup$ I'm sorry, but I still cannot understand why it has to be in Lie algebra and couldn't be left-translated element of Lie algebra for example? $\endgroup$
    – Darkseid
    Commented Jul 27, 2019 at 13:47
  • $\begingroup$ Well the partial derivative is some sort of "infinitesimal difference quotient." To make it covariant, we add on a rotation to it--an "infinitesimal rotation." Really small rotations should be really close to $1_G \in G$, i.e. in the tangent space to the identity element. (Sort of a hand-wavey response, I realize) $\endgroup$
    – Todd N
    Commented Jul 28, 2019 at 8:57

I think I got it sorted out, at least motivationally.

Apart from the usual definition of a matrix of 1-forms with peculiar transformation properties, the connection could also be defined through identification of fibers at different points of base manifold. Unsurprisingly, it would make a lot of sense if these fibers are identified through multiplication by an element of Lie Group $G$. Now for the infinitesimal form of such identification we would naturally get Lie Algebra $\mathfrak{g}$.

Note that it is by no means required that the connection takes values in $\mathfrak{g}$ (unless we deal with Principal Bundle). The argument above might be considered an inspiration at best, not a mathematical treatment of the question.


Let the structure group be $\operatorname{O}(m),\operatorname{U}(m), \operatorname{SO}(m),\operatorname{SU}(m)$ (or $\operatorname{O}(k,l),...$), and the parallel transport preserve metric of Riemannian/Hermitian vector bundle $\sqcup_{x\in M} V_x$. $\newcommand{\op}[1]{\operatorname{#1}} \newcommand{\map}[5]{\begin{array}{cccc} #1 & #2 & \longrightarrow & #3 \\ & #4 & \longmapsto & #5 \end{array}} \newcommand{\K}{\mathbb{K}} \newcommand{\R}{\mathbb{R}} \newcommand{udb}[2]{\underbrace{#1}_{#2}} \newcommand{\e}[1]{\frac{\partial}{\partial #1}} \newcommand{\al}[1]{\begin{align*} #1 \end{align*}} \newcommand{\br}[1]{\left(#1\right)}$

Let $\gamma:[0,a]\to M$ be a smooth curve with initial value $\gamma(0)=p,\ \dot\gamma(0)=X$; Denote the parallel transport $P(\gamma)_t^0: V_{\gamma(t)}\to V_{\gamma(0)}$.

The definition of $G$ structure bundle chart ($G=\operatorname{SO}(m),...$) require that, the bundle chart vector frame $\phi^{-1}(x,e_j)$ is orthonormal. On Riemannian/Hermitian bundle, local $C^\infty$ orthonormal frame can be constructd by Gram-Schmidt procedure (So if we have this metric, we have this structure bundle chart), see https://math.stackexchange.com/a/3002983/634573. So, let $E_j|_x\in \sqcup_{x\in W} V_x$ be a local $C^\infty$ orthonormal frame around $p$. Compute the covariant derivative

$$ \nabla_X E_j=\lim_{t\to 0}\frac{P(\gamma)_t^0 E_j|_{\gamma(t)}-E_j|_{\gamma(0)}}{t}=A_X(p)_{kj} E_k|_p $$

By metric preserving, $P(\gamma)_t^0 E_j|_{\gamma(t)}$ is another orthonormal frame of $V_p$, so there is curve $t\xrightarrow{C^\infty} U(t)\in \operatorname{SO}(m),\cdots$ with $U(0)=I$, such that $$ P(\gamma)_t^0 E_j|_{\gamma(t)}= U(t)_{kj} E_k|_{p}=(E_1|_p,...,E_1|_p)U(t)\left(\array{ \vdots \\ \underbrace{1}_{\text{$j$-th}} \\ \vdots }\right) $$ Use notation $(w_1,...,w_n)=(v_1,..,v_n)\cdot \left(\array{ a_{11} &\cdots & a_{1n} \\ \vdots & \ddots &\vdots \\ a_{n1} & \cdots & a_{nn}}\right)=(a_{k1} v_k ,...,a_{kn} v_k )$. Now we have $$ \nabla_X E_j=(E_1|_p,...,E_1|_p)\cdot \underbrace{\lim_{t\to 0}\frac{U(t)-I}{t}}_{=\dot U(0)}\cdot\left(\array{ \vdots \\ \underbrace{1}_{\text{$j$-th}} \\ \vdots }\right)=\dot U(0)_{kj} E_k|_p $$ So we get Lie algebra valued $A_X(p)=\dot U(0)\in \mathsf{so}(m),\mathsf{su}(m),...$

Let $(F_1,...,F_m)$ be another local $C^\infty$ frame, and $B_X(x)$ the corresponding connection matrix; Let $(F_1,...,F_m)|_x=(E_1,...,E_m)|_x\cdot T(x)$. Then the the transformation rule:

$$ B=T^{-1}\cdot dT + T^{-1}\cdot A\cdot T $$

The proof is: write

$$(\nabla_X E^1,\ldots,\nabla_X E^m) (p)= (E^1,...,E^m)|_p\cdot A_X(p)$$

Let $(F^1,...,F^m)|_x=(E^1,...,E^m)|_x\cdot T(x)$. It can prove that

$$ \al{(\nabla_X F^1,..., \nabla_X F^m)&=(E^1,...,E^m)\cdot \partial_X T + (\nabla_X E_1,...,\nabla_XE^m)\cdot T \\ &= (E^1,...,E^m) \br{ \partial_X T + A_X\cdot T}} $$

On the other hand,

$$\al{(\nabla_X F^1, ..., \nabla_X F^m)&=(F^1, ..., F^m)\cdot \tilde A_X \\ &=(E^1,..., E^m)\cdot T\cdot \tilde A_X }$$

So $\tilde A_X=T^{-1}\cdot\partial_X T +T^{-1} A_X \cdot T$. Or proof via computation: $$ \al{\nabla_{X} \underbrace{F_j}_{\sum_{\alpha=1}^m T_{\alpha j} E_\alpha}&=\sum_{\alpha=1}^m\partial_X T_{\alpha j}E_{\alpha}+ \sum_{\alpha=1}^mT_{\alpha j} \underbrace{\nabla_X E_{\alpha}}_{\sum_{\beta=1}^m(A_X)_{\beta\alpha} E_\beta} \\ &= \sum_{\beta=1}^m\left(\partial_X T_{\beta j}+ \sum_{\alpha=1}^m T_{\alpha j} (A_X)_{\beta \alpha}\right)\underbrace{E_\beta}_{\sum_{k=1}^m(T^{-1})_{k\beta} F_k}} $$ Then

$$ \al{\underbrace{(B_X)_{kj}}_{\text{$k$ row, $j$ colume of $B_X$}}&=\underbrace{\sum_{\beta=1}^m (\partial_X T_{\beta j}) (T^{-1})_{k\beta}}_{\text{$k$ row, $j$ colume of $T^{-1}\cdot(\partial_X T)$}}+\underbrace{\sum_{\alpha,\beta=1}^m T_{\alpha j} (A_X)_{\beta\alpha} (T^{-1})_{k\beta}}_{\text{$k$ row, $j$ colume of $T^{-1}\cdot A_X\cdot T$}} \\ B_X&=T^{-1}\cdot(\partial_X T)+T^{-1}\cdot A_X\cdot T} $$

If $F_i$ is also an orthonormal frame, or equivalent $T(x)\in \operatorname{SO}(m),...$ then we also have $B_X(p)\in \mathsf{so}(m),...$ On the other hand, $$ T^{-1}\cdot(\partial_X T) +T^{-1}\cdot A_X\cdot T=\frac{d}{dt}|_{t=0} \left(T(p)^{-1}\cdot T\big(\gamma(t)\big) + T(p)^{-1}\cdot U(t)\cdot T(p) \right) $$

(1) If [structure group is matrix group $G\subset \operatorname{GL}(m,\mathbb K)$] require that [we select a family of basis $\{(E_1,...,E_m)|_x\cdot T\}_{x\in W \\ T\in G}$ of every fiber $V_p$], where $E_j|_x=\phi^{-1}(x,e_j)$ is bundle chart vector basis.

Another bundle chart vector basis $F_j|_x=\psi^{-1}(x,e_j)$ is also in this family, $$ \require{AMScd} \begin{CD} V_x @>{\text{identity map, $(F_1,...,F_m)|_x=(E_1,...,E_m)|_x\cdot g_{\psi\phi}(x)$}}>> V_x \\ @V{(W_\psi,\psi)}VV@VV{(W_\phi,\phi)}V\\ \mathbb K^n @>>{g_{\psi\phi}(x)}> \mathbb K^n \end{CD} $$

(2) Once again, assume that the parallel transport (or say it connection) maps between these basis, that is, $\{P(\gamma)_t^0 E_j|_{\gamma(t)}\}_{j=1}^m\in \{(E_1,...,E_m)|_{\gamma(0)}\cdot T\}_{T\in G}$. Or in other words, as OP's self answer say, we require that, the bundle chart representation of parallel transport $P(\gamma)_t^0:V_{\gamma(t)}\to V_{\gamma(0)}$ is an element of $G\subset \operatorname{GL}(m,\mathbb K)$.

By same argument, there is $t\xrightarrow{C^\infty} U(t)\in G$ with $U(0)=I$ such that $$ P(\gamma)_t^0 E_j|_{\gamma(t)}=(E_1|_p,...,E_1|_p)U(t)\left(\array{ \vdots \\ \underbrace{1}_{\text{$j$-th}} \\ \vdots }\right) $$ Then similarly we can get $\nabla_X E_j=\dot U(0)_{kj} E_k|_p$ and Lie algebra valued $A_X(p)=\dot U(0)\in \mathsf{g}$.

The converse is also ture. That is, if the connection matrix is Lie algebra valued, then connection preserve $G$ structure.

Let $\forall_X$ connection matrix $A_X$ be Lie algebra valued. Let $P(\gamma)_0^t E_\mu|_p= Z_\mu{}^k(t) E_k|_{\gamma(t)}$. Compute parallel transport equation $$ 0=\nabla_{\dot\gamma(t)} (Z_\mu{}^k(t)E|_{\gamma(t)})=\bigg(\dot Z_\mu{}^k(t)+\udb{\dot\gamma^i A_{i}}{=A_{\dot\gamma(t)}} \big(\gamma(t )\big)^k{}_{j} Z_\mu{}^j(t)\bigg) E_k|_{\gamma(t) } $$ We obtain the system of ODE equations $$ \al{\dot Z_\mu{}^1&= -A_{\dot\gamma(t)}\big(\gamma(t)\big)^1{}_j Z_\mu{}^j\\ &\cdots\\ \dot Z_\mu{}^m&= -A_{\dot\gamma(t)}\big(\gamma(t)\big)^m{}_j Z_\mu{}^j} \\ \text{initial value:}\ Z_\mu{}^k(0)= \delta_\mu{}^k $$ Written in vector form this is $$ \dot Z_\mu(t)=-A_{\dot \gamma(t)}\big(\gamma(t))\big) Z_\mu(t),\quad Z_\mu(0)=\br{\array{ \vdots \\ \underbrace{1}_{\text{$\mu$-th}} \\ \vdots }} $$ Write $Z_1,\ldots,Z_m$ in matrix form, that is $$ \dot Z(t)= -\big(A_{\dot\gamma(t)}\circ \gamma\big)(t)\cdot Z(t),\quad Z(0)=I $$ Because for each $t$, the matrix $A_{\dot \gamma(t)}\in \mathsf g$ is Lie algebra-valued. Then how to prove that $Z(t)\in G$? Let us abbreviate $-(A_{\dot{\gamma}(t)}\circ\gamma)(t)$ to $A(t)$.

We find that this ODE is a time-varying vector field on $\op{Matrix}(m,\K)\cong \K^{m^2}$ $$ \map{X:}{\R\times \K^{m^2}}{\K^{m^2}}{(t,B)}{A(t)B} $$ of the integral curve.

If we consider $M=\op{Matrix}(m,\K)$ as a real manifold (notation $M$ dose not mean the original base manifold), then the value domain should be $\sqcup_{B\in M}T_B M$. The matrix Lie group $G$ is a real sub-manifold of $M$. However, it’s troublesome to convert complex matrix to real matrix, via some linear isomorphism induced by $\op{GL}(m,\mathbb C),\op{GL}(2m,\R)$, and $(x_1+iy_1,...,x_m+iy_m)\longleftrightarrow \left\{\array{(x_1,y_1,...,x_m,y_m) \\ (x_1,...,x_m,y_1,...,y_m)}\right.$. When we write $\lim_{t\to 0}\frac{U(t)-I}{t}$ to get the tangent vector, we want this compatible in both real and complex case. An observation is that, $\frac{U(t)-I}{t}$ just use linearity, but not complex multiplication.

If $B\in G$, then $A(t)\cdot B\in T_{B} G$. This can be verified via curve: since $A(t)\in \mathsf g$ is Lie algebra valued, there exists a curve $s\mapsto \alpha(s)\in G$ such that $\alpha(0)=I,\ \dot \alpha(0)=A(t)$.

Then for the curve $s\mapsto \alpha(s) B\in G$, we have $$ \alpha(0)B=B\\\frac{d}{ds}\Big|_{s=0} \alpha(s)B=\udb{\dot\alpha (0)}{A(t)}B $$ Thus we get $A(t)B \in T_B G$.

Now the problem translates to: if $$ \map{X:}{\R\times M}{\sqcup_{p\in M}T_p M}{(t,p)}{X(t,p)} $$ is a time-varying tangent vector field on the $N$ dimension real manifold, and assume that, when $X$ is restricted to the sub-manifold $S$, it becomes a vector field on $S$. If the initial value of the integral curve $\phi(t)$ is also on $\phi(0)\in S$, does the whole integral curve stay inside the sub-manifold $S$ and not go out of it?

Recall that, the ODE on the (real) manifold is indeed well-defined and does not depend on the choice of the coordinate chart. The solution of the vector field $X$ — i.e. the integral curve –- is the map $\map{\phi:}{\R}{M}{t}{\phi(t)}$ satisfying $\frac{d}{dt} \phi(t)=X(t,\phi(t))$. When write down this equation, it already does not depend on the coordinates chart. But we still need to look at, how it look like in the coordinate chart $(U,x^1,... ,x^m)$. Since $\phi(t)=\phi^k(t)\e{x^k}$ and $X(t,p)=X^k(t,p)\e{x^k}$. The ODE should be $$ \frac{d}{dt}\phi^k(t)=X^k(t,\phi^1(t),... ,\phi^m(t)) $$ Both sides are contravariant in $\R^N$. Thus, for the case of sub-manifolds, we can choose a coordinate chart such that $x|_{S\cap U}(p)=(x^1,...,x^l ,0,... ,0)(p)$. Together with the assumption that $X|_{S}(t,p)\in T_p S$, we also have $X|_{S\cap U}(t,p)=(X^1,... ,X^l,0,... ,0)(t,p)$.

So the ODE in the coordinate chart can be solved first for the part of $\K^l$, and then construct a solution on $\K^m$ — let the $(l+1)\sim m$ component be zero. Then, by the uniqueness of the ODE, the integral curve will be restricted to $\K^l$ in the coordinate system. That is, the integral curve is restricted to the sub-manifold $S$.

Applying to our case $M=\op{Matrix}(m,\K)\cong \K^{m^2}$ and $S=G$, the ODE $\dot Z(t)=A(t)Z(t),\ Z(0)=I$, we get, the integral curve $Z(t)\in G$.


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