I would like to confirm that the Hessian of a smooth function $f\colon U\to \mathbb R$ that is defined on an open subset $U\subset M$ of a smooth manifold $M$ is in local coordinates given by $$\nabla^2 f = \sum_{i,j} \frac{\partial^2 f}{\partial x_i\partial x_j} dx_i\otimes dx_j – \sum_{j,k}\frac{\partial}{\partial x_j}\omega_{jk}\otimes dx_k$$
Unfortunately, I am still having a hard time doing these kind of problems. I really hope someone could give me a hint.
What I know:
I know that $$\nabla^2_{X,Y} f = XYf – (\nabla_XY)f$$ $$\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j} = \sum_k\Gamma^k_{ij}\frac{\partial}{\partial x_k}$$ and $$\nabla\frac{\partial}{\partial x_j} = \sum_k\omega_{kj}\otimes
\frac{\partial}{\partial x_k}$$
What I tried:
So I've tried expressing $X$ resp. $Y$ in terms of local coordinates s.t. $$X= \sum_i X_i\frac{\partial}{\partial x_i},\ Y= \sum_j Y_j\frac{\partial}{\partial x_j}$$ and computing $\nabla^2_{X,Y}f$ accordingly but that didn't seem to be what I'm supposed to do.
Then I tried writing $XYf = X(Yf) = X(\nabla_Yf)$ with respect to $X = \frac{\partial}{\partial x_i}, Y =\frac{\partial}{\partial x_j}$ but that approach wasn't succesfull either.
I had the impression that the most reasonable approach was to express $df = \sum_j \frac{\partial f}{\partial x_j}dx_j$ and then compute
\begin{align}
\nabla df = \nabla\left(\sum_j \frac{\partial f}{\partial x_j}dx_j\right)
&= \sum_j \nabla \left(\frac{\partial f}{\partial x_j}dx_j\right) \\ &= \sum_j \left(\nabla\frac{\partial f}{\partial x_j}\otimes dx_j + \frac{\partial f}{\partial x_j}\nabla dx_j\right) \\
&= \sum_j \left(\sum_i\omega_{ij}\otimes \frac{\partial f}{\partial x_i} \otimes dx_j + \frac{\partial f}{\partial x_j}\nabla dx_j\right) \\
&= \sum_j \left(\sum_i\omega_{ij}\otimes \frac{\partial f}{\partial x_i} \otimes dx_j – \frac{\partial f}{\partial x_j}\sum_k \omega_{kj}\otimes dx_k \right) \\
&= \sum_j \left(\sum_i\omega_{ij}\otimes \frac{\partial f}{\partial x_i} \otimes dx_j – \frac{\partial f}{\partial x_j}\sum_k \omega_{kj}\otimes dx_k\right) \\
&= \sum_{i,j} \omega_{ij}\otimes \frac{\partial f}{\partial x_i} \otimes dx_j – \sum_{j,k}\frac{\partial f}{\partial x_j} \omega_{kj}\otimes dx_k
\end{align}
Edit:
I've added more steps and it seems that I'm quite close. The problematic part here is the left hand side. I think the only choice I have is trying to express $$\omega_{kj} = \sum_i\Gamma^k_{ij}dx_i$$ but how do get to $\frac{\partial^2f}{\partial x_i\partial x_j}$?
Any help is highly appreciated!
Best Answer
Your expression is wrong. There are no derivatives of $\omega$ involved (also notice that there is no first derivative of $f$ in the last parcel of your formula). Recall that for a smooth local frame $\{e_1, \cdots, e_n\}$ the connection forms $\{\omega_{j}^{k} \}_{1 \leq k \leq n}$ are determined by $$\nabla_{e_i} e_j = \sum_{1 \leq k \leq n} \omega_{j}^{k}(e_i) e_k$$
Or, equivalently, $$\nabla e_j = \sum_{1 \leq k \leq n} \omega_{j}^{k}(\bullet) e_k$$
which is usually written as:
$$\nabla e_j = \sum_{1 \leq k \leq n} \omega_{j}^{k} e_k$$
In the case that the local frame is a coordinate frame (which is apparently the case you want), by definition $$\omega_j^k(\partial_{i}) = \Gamma_{ij}^k$$ Now, since $\nabla^2 f$ is a $(0, 2)$ tensor and $\{ \mathrm{d}x^i \otimes \mathrm{d}x^j \} $ forms a basis of the space of such tensors, we have:
$$\nabla^2 f = \sum_{1 \leq i, j \leq n} (\nabla^2 f)_{ij} \mathrm{d}x^i \otimes \mathrm{d}x^j $$
But
\begin{aligned} (\nabla^2 f)_{ij} = \nabla^2_{\partial_i, \partial_j} f &= \nabla_{\partial_i} \nabla_{\partial_j} f - (\nabla_{\partial_i} \partial_j)f \\ &= \frac{\partial^2 f}{\partial x^i \partial x^j} - \sum_{1 \leq k \leq n} (\Gamma_{ij}^k \partial_k) f \\ &= \frac{\partial^2 f}{\partial x^i \partial x^j} - \sum_{1 \leq k \leq n} \Gamma_{ij}^k \frac{\partial f}{\partial x^k}\\ \end{aligned}
Which can also be written as
$$(\nabla^2 f)_{ij} = \frac{\partial^2 f}{\partial x^i \partial x^j} - \sum_{1 \leq k \leq n} \omega_{j}^k(\partial_i) \frac{\partial f}{\partial x^k}$$
So we have:
$$\nabla^2 f = \sum_{1 \leq i, j \leq n }\left( \frac{\partial^2 f}{\partial x^i \partial x^j}- \sum_{1 \leq k \leq n} \omega_{j}^k(\partial_i) \frac{\partial f}{\partial x^k}\right) \mathrm{d}x^i \otimes \mathrm{d}x^j $$
Clearly,
$$\sum_{1 \leq i \leq n} \omega_{j}^k(\partial_i) \mathrm{d} x^i = \omega_{j}^k$$
Therefore:
$$\nabla^2 f = \left( \sum_{1 \leq i, j \leq n } \frac{\partial^2 f}{\partial x^i \partial x^j} \mathrm{d}x^i \otimes \mathrm{d}x^j \right) - \sum_{1 \leq j, k \leq n} \frac{\partial f }{\partial x^k} \omega_{j}^k \otimes \mathrm{d}x^j$$
which can also be written as
$$\nabla^2 f = \left( \sum_{1 \leq i, j \leq n } \frac{\partial^2 f}{\partial x^i \partial x^j} \mathrm{d}x^i \otimes \mathrm{d}x^j \right) - \sum_{1 \leq j, k \leq n} \left(\frac{\partial }{\partial x^k} \otimes \omega_{j}^k \right)(f, \bullet) \otimes \mathrm{d}x^j$$
REMARK: A vector field $X$ can be seen as a $(1, 0)$ tensor field, which acts on the space of $1$-forms $\Gamma(T^{*} M)$ by $X(\omega) = \omega(X)$. $\nabla X$ can be seen as a $(1, 1)$ tensor field: either by seeing it as a map which takes a vector field as input and outputs another vector field, os a map which takes as input a real smooth function and another vector field and outputs a real smooth function, that is: \begin{aligned} \nabla X: \Gamma(T^{*} M) \times \Gamma(T M) &\to \mathcal{C}^{\infty}(M) \\ (\beta, Y) &\mapsto (\nabla_Y X)(\beta) = \beta(\nabla_Y X) \end{aligned}
Now, if $V$ is a vector field and $\beta$ is a $1$-form, we have the tensor product $\beta \otimes V: \Gamma(T M) \times \Gamma(T^{*} M) \to \mathcal{C}^{\infty}(M)$, which is a tensor field of type $(1, 1)$ given by $$(\beta \otimes V)(Z, \alpha) = \beta(Z)V(\alpha)= \beta(Z) \alpha(V)$$
In particular, $\nabla e_i$ is a $(1, 1)$ tensor field given by $$(\nabla e_i)(X, \alpha) = (\nabla_X e_i)(\alpha) = \sum_{1 \leq j \leq n} (\omega_i^j(X) e_j)(\alpha) = \sum_{1 \leq j \leq n} \omega_i^j(X) e_j(\alpha) = \sum_{1 \leq j \leq n} (\omega_i^j \otimes e_j)(X, \alpha)$$
Since $X$ and $\alpha$ are arbitrary, we can write:
$$\nabla e_i = \sum_{1 \leq j \leq n} \omega_i^j \otimes e_j$$