2

I have been reading Chapter 0 of Griffiths' and Harris' Principles of Algebraic Geometry, in particular, the section on Vector Bundles, Connections, and Curvature. I have three questions:

A connection is defined on page 72 as follows: A connection $D$ on a complex vector bundle $E \to M$ is a map $$D : \Omega^0(E) \longrightarrow \Omega^1(E)$$ satisfying the Leibniz rule $$D(f \cdot \zeta) = df \otimes \zeta + f \cdot D(\zeta),$$ for all sections $\zeta \in \Omega^0(E)(U)$, $f \in \mathscr{C}^{\infty}(U)$.

Q1. Why does a connection take values in the space of smooth 1-forms on $E$? A connection yields a way of differentiating sections, generalising the directional derivative on functions. We remind ourselves that the directional derivative of a function $f \in \mathscr{C}^{\infty}(M)$ in the direction of $v \in T_p M$ is the smooth function $D_v f \in \mathscr{C}^{\infty}(M)$ given by $\nabla f \cdot v$.

Is $\nabla f \cdot v$ a 1-form? -- I am not seeing the compatibility here.

Q2. I have previously seen the covariant derivative be defined as a map $D : \Gamma(E) \longrightarrow \Gamma(E) \otimes \Gamma(T^{\ast} M)$ satisfying $D_{v+u}(\sigma) = D_v(\sigma) + D_u(\sigma)$, $D_{f v} (\sigma) = f D_v(\sigma)$, $\mathbb{R}$-linearity in $\sigma$, and the Leiniz rule. Do these definitions coincide, i.e., do we get tensoriality, $\mathbb{R}$-linearity, etc., from the Leibniz rule alone?

Q3. What is the distinction between a unitary frame and a holomorphic frame? Griffiths and Harris do not seem to define the distinction, but they are distinguished (see, e.g., page 73).

Please let me know if anything is not clear, or whether these questions have already been asked here. Thanks in advance.

AmorFati
  • 1,888
  • In question 1, what do you mean by $\nabla f\cdot v$? Are you restricting your attention to the case of $M = \mathbb{R}^n$ where by $\nabla f$ you mean $\left(\frac{\partial f}{\partial x_1}, \dots, \frac{\partial f}{\partial x_n}\right)$? – Michael Albanese Dec 18 '18 at 02:27
  • @MichaelAlbanese Yes, that is correct, apologies for not being precise, or even defining the notation that I have used. – AmorFati Dec 18 '18 at 02:42

1 Answers1

1

  1. Recall that a one-form can be viewed as a $\mathcal{C}^{\infty}(M)$-linear map $\Gamma(TM) \to \mathcal{C}^{\infty}(M)$. Given a smooth function $f$ on $\mathbb{R}^n$ and a smooth vector field $V \in \Gamma(T\mathbb{R}^n)$, the map $V \mapsto \nabla f\cdot V$ is $\mathcal{C}^{\infty}(\mathbb{R}^n)$-linear and hence defines a one-form on $\mathbb{R}^n$.

  2. Note that $\Omega^1(E) = \Gamma(TM\otimes E) \cong \Gamma(TM)\otimes\Gamma(E)$; for the last isomorphism, see this question. As for linearity, one needs to add that to the definition because it does not follow from the Leibniz rule.

  3. A unitary frame for a hermitian metric $h$ is a frame $\{\sigma_1, \dots, \sigma_n\}$ such that $h(\sigma_i, \sigma_j) = \delta_{ij}$. On the other hand, a holomorphic frame is just a frame of holomorphic sections of the holomorphic tangent bundle.

Michael Albanese
  • 99,526