3
$\begingroup$

Let $L \to M$ be a complex line bundle over a smooth manifold $M$. Let $\{U_\alpha\}$ be a trivializing open cover with transition maps $z_{\alpha\beta}:U_\alpha \cap U_\beta \to \Bbb C^* = \text{GL}(1,\Bbb C)$. The bundle of endomorphisms of $L$, $\text{End}(L) \cong L^* \otimes L$ is trivial since it can be defined by transition maps $(z_{\alpha\beta})^{-1} \otimes z_{\alpha\beta} = 1$. This, the space of connections on $L$, $\mathcal{A}(L)$ is an affine space modeled by the linear space of complex valued $1$-forms. A connection on $L$ is simply a collection of $\Bbb C$-valued $1$-forms $\omega^\alpha$ on $U_\alpha$ related on overlaps by $$\omega^\beta = \frac{dz_{\alpha\beta}}{z_{\alpha\beta}} + \omega^\alpha = d\log z_{\alpha\beta} + \omega^\alpha.$$

Could someone explain to me what is going on here? I'm familiar with line bundles, transition maps, connections and bunch of other stuff, but this seems very weird to me.

I have never seen for example this affine space $\mathcal{A}(L)$ which I believe is defined to be somehow realted to $\Omega^1(\text{End}(L))$. This expression involving the logarithm is also pulled out of thin air, how does one go about deriving such a thing?

$\endgroup$

2 Answers 2

7
$\begingroup$

Let $\nabla_0\colon C^\infty(M,L)\to \Omega^1(M,L)$ be a reference connection. Then any other connection on $L$ can be written as $\nabla_A = \nabla_0+A$, where $A\in \Omega^1(M,\mathbb C)\cong \Omega^1(M,\mathrm{End}(L))$. This is what makes $\mathcal{A}(L)$ into an affine space.

Let now $(U_\alpha)$ be an open cover of $M$ and $s_\alpha\in C^\infty(U_\alpha,L|_{U_\alpha})$ be collection of non-vanishing sections. These automatically trivialise the bundle and thus there exist transition functions $z_{\alpha\beta}\in C^\infty(U_{\alpha\beta},\mathbb C)$ on $U_{\alpha\beta} =U_\alpha\cap U_\beta$ such that $$s_\alpha = z_{\alpha\beta}s_\beta \quad \text{ on } U_{\alpha\beta}. \tag{1}$$ Moreover, we can express $\nabla_A s_\alpha$ via some $\omega_{\alpha}\in \Omega^1(U_\alpha,\mathbb C)$ as follows: $$ \nabla_A s_\alpha = \omega_{\alpha} s_\alpha \tag{2} $$ Now apply $\nabla_A$ to $(1)$ and use the Leibniz rule: $$ \omega_\alpha s_\alpha = \nabla_A(s_\alpha) = dz_{\alpha\beta} s_\beta + z_{\alpha\beta} \nabla_As_{\beta} = dz_{\alpha\beta} s_\beta + z_{\alpha\beta} \omega_\beta s_\beta = \frac{dz_{\alpha\beta}}{z_{\alpha\beta}} s_\alpha + \omega_\beta s_\alpha $$ Since the $s_\alpha'$s don't vanish we get the following relation you encountered: $$ \omega_\alpha = \frac{dz_{\alpha\beta}}{z_{\alpha\beta}} + \omega_\beta \tag{3} $$ What I've explained is how to associate to a connection $\nabla_A$ a collection of $1$-forms $(\omega_\alpha)$ satisfying relation $(3)$. Now try to reverse engineer a connection $\nabla_A$ out of the $1$-forms!

$\endgroup$
7
  • $\begingroup$ Is this kinda the same thing as in the real case, given frames $e_1,\dots, e_r$ of a rank $r$ bundle $E$ we have $$\nabla_X e_j = \sum \omega^i_j(X)e_i$$ for some $1$-forms $\omega^i_j$ and a vector field $X$? @jan-bohr $\endgroup$
    – Tepes
    Mar 1 at 16:39
  • $\begingroup$ That's true, though you might want to write $e_{\alpha,j}$ and $\omega_{\alpha,j}^i$ to match the notation above. For a $\mathbb K$ vector bundle of rank $r$, you'll then have $\omega_\alpha\in\Omega^1(U_\alpha,\mathbb K^{r\times r})$, for both $\mathbb K=\mathbb R$ and $\mathbb K=\mathbb C$. $\endgroup$
    – Jan Bohr
    Mar 1 at 16:56
  • $\begingroup$ Is there an implicit sum in $\nabla_A s_\alpha = \omega_{\alpha} s_\alpha$? I'm wondering if this is the same as $De_j = \displaystyle\sum_i \omega_{ji} \otimes e_i $ what Azur wrote @jan-bohr. $\endgroup$
    – Tepes
    Mar 2 at 10:27
  • $\begingroup$ No, in the case of line bundles (i.e. $r=1$ and hence $i=j=1$) there is no implicit sum. The $\alpha$-index only indicates which patch you are on. $\endgroup$
    – Jan Bohr
    Mar 2 at 12:03
  • 1
    $\begingroup$ Yes I suppressed the $\otimes$-sign. Also, my choice of putting $\alpha$ into the lower index was arbitrary. $\endgroup$
    – Jan Bohr
    Mar 3 at 9:18
3
$\begingroup$

Jan Bohr's answer is already a great answer on how to solve your problem, let me just add a bit more on that. In particular the intuition of where the logarithm comes from.

Most of this answer will actually just be re-stating definitions that I'll need, so you can just skip to the last part to see the explanation.. This is all inspired from Huybrechts's 2005 book on complex geometry, which I'll write [1] for short:


  • Let $M$ be a complex manifold and $E \to M$ a bundle over it (for now, of any rank), then a connection is defined the usual way:

\begin{equation} D : \Gamma(E) \longrightarrow \Omega_\mathbb{C}^1(E) := T^{*}M\otimes\Gamma(E) \end{equation}

In other word, given $\sigma \in \Gamma(E)$, and $V \in TM$, $D\sigma(V) = D_V \sigma$ is a section in $\Gamma(E)$. And we ask $D$ to satisfy the usual properties a connection satisfies. Then:

  • If we let $(\cdot, \cdot)$ be a Hermitian structure on $E$, then we say that $D$ is compatible with the Hermitian structure if $D(s, r) = (Ds, r) + (s, Dr)$.

On vector bundles (where we'll consider the example $TM$), there is actually more structure:

  • Recall that since $M$ is complex, $TM$ splits into $TM^{1,0} \oplus TM^{0,1}$, and so $\nabla$ descends to a connection $\nabla^{0,1}$ on $TM^{0,1}$. (Lemma 4.2.7 of [1]. This is a general fact on connections on split bundles.)
  • We say that $\nabla$ is compatible with the holomorphic structure if $\nabla^{0,1} = \bar{\partial}$, the CR-operator.
  • Then, Corollary 4.2.13 of [1] tells you that the space of connections $\mathcal{A}(L)$ is an affine space over $\Omega_\mathbb{C}^{1,0}(M, \text{End}(E))$ (with slightly different notation in the book).

Note: actually, I realized Jan Bohr had already explained how to do that at the start of their answer, and that I'd missed it.


We need a tiny bit more background on Chern connections before we derive this logarithm thing:

If you choose a holomorphic frame $e_1, ..., e_r$ on the bundle $E$, then you can find $1$-forms $\omega_{ji} \in \Omega_\mathbb{C}^1(M)$ such that:

\begin{equation} De_j = \displaystyle\sum_i \omega_{ji} \otimes e_i \end{equation}

(These are the same ones that Jan Bohr talks about in their answer).

Now, if we assume that $D$ is compatible with the holomorphic structure (a Chern connection), then one can show that all these $1$-forms are holomorphic (let me know if you'd like me to add a proof of this).


Finally, let us restrict our attention to a holomorphic line bundle $E$, with a hermitian metric $(\cdot,\cdot)$, and a Chern connection (compatible with both $(\cdot, \cdot)$ and the complex structure). Then, there is only one $1$-form, $\omega$, and the holomorphic frame is simply a section $e : M \to TM^{1,0}$. The hermitian metric is then entirely determined by the following function:

\begin{equation} h = (e,e) : M \longrightarrow \mathbb{C} \end{equation}

Observe that since $e$ is holomorphic, $h$ is as well, and so $\bar{\partial} h = 0$.

Because of the compatibility condition between $D$ and $(\cdot, \cdot)$, we have:

\begin{align} \mathrm{d}h &= (De, e) + (e, De) \\ &= (\omega \otimes e, e) + (e, \omega \otimes e) \\ &= \omega h + \bar{\omega} h \end{align}

(To explain the last term: by linearity, you can move the tensor product out of the Hermitian brackets, so $(\omega\otimes e, e) = \omega (e,e) = \omega h$. The other term is then simply its conjugate, so $(e, \omega\otimes e) = \overline{\omega h} = \bar{\omega} h$).

But then, by definition of $\partial$ and $\bar{\partial}$, you also have $\mathrm{d} = \partial + \bar{\partial}$. You can hence identify:

\begin{equation} \begin{cases} \partial h = \omega h \\ \bar{\partial} h = \bar{\omega} h \end{cases} \end{equation}

(This identification uses the fact that the $1$-form $\omega$ is holomorphic).

Now, notice that these two equations are conjugate, so there is really only one, $\partial h = \omega h$. This is an equality on 1-forms; and therefore, the way we write the right-hand side is a bit confusing. We should rather write $h \omega$, since the function $h$ is simply a factor in front of the $1$-form $\omega$. We can therefore rewrite:

\begin{equation} \omega = h^{-1} \partial h \end{equation}

And, as you can notice, the right-hand side very much looks like $\partial \text{log} h$, hence why we write $\boxed{\omega = \partial \text{log}h}$.

Note: Now, to actually be able to write this equality rigorously on $E$, we'd need to check that $\text{log} h$ is well-defined on $E$, that there's no branch cuts,... And I admit that is beyond my knowledge in complex geometry. However, this is very doable if you work in local coordinates, which is exactly the content of Jan Bohr's comment (where he also explains how to get the family $\omega_{ij}$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .