Understanding Hessian on Manifold (without Riemannian Geometry) I've been going through notes on Morse theory and Handlebody theory and I've been having some trouble with the definition of the Hessian provided. The notes are on pages 3-4 here http://people.math.gatech.edu/~etnyre/preprints/papers/PCMI-HandlebodyA-B.pdf
For the setup, we have a manifold $W$, a point $p$ and a map $f : W \to \mathbb{R}$ that is Morse. For my current definition of being Morse, it means that at any critical point, $df$ is transverse to the zero section $Z$ at $p$. Considering $df$ as a map from $W \to T^*W$, we can consider the derivative of this map at the point $p$, i.e. $d_p(df):T_pW \to T_{df(p)}(T^*W))$. Since $p$ is a critical point, $df(p) = (p,0)$. So $d_p(df): T_pW \to T_{(p,0)}(T^*W)$. We also have a splitting $T_{(p,0)}(T^* W) \cong T_{(p,0)}(Z) \oplus T_p^*M$ which I am also comfortable with. This gives us a projection map $C_p: T_{(p,0)}(T^* W) \to T_p^*M$. Finally, the Hessian at the point $p$ is defined as $(d^2f)_p : T_pW \times T_pW \to \mathbb{R}$ by $(v,w) \mapsto C_p(d_p(df(v)))(w)$. I sort of understand what's going on but not really, and I'm not able to show that this is symmetric (though the bilinearity is obvious I think).
I want to show that $C_p(d_p(df(v)))(w) = C_p(d_p(df(w)))(v)$. 
My first thought is to use local coordinates. Let $U$ be a coordinate patch for $W$, and so we have $f: U \to \mathbb{R}$. In local coordinates, $df(p) = (p, \sum_i \frac{\partial f}{\partial x^i} dx^i)$, as a map from $W \to T^* W$. Now I'm going to try and define $d_p(df):T_pW \to T_{(p,0)} (T^* W)$. This just differentiates with respect to coordinates on $W$, so I get $(\sum_i \frac{\partial p}{\partial x^i} dx^i, \sum_j \sum_i \frac{\partial^2}{\partial x^i \partial x^j} dx^i dx^j)$.  This is where I start to get really confused. Is there supposed to be a wedge between $dx^i$ and $dx^j$? I feel like there shouldn't, but isn't that expression meaningless without a wedge? Is it supposed to be $dx^i \otimes dx^j$? Anyway, moving onto projecting using $C_p$ gives me  $C_p(d_p)(df) = \sum_j \sum_i \frac{\partial^2}{\partial x^i \partial x^j} dx^i dx^j$ which indeed does look like the Hessian in local coordinates. I know the definition wants me to contract this with $v = \sum v^i \frac{\partial}{\partial x^i}$ to get a covector and then apply to $w = \sum w^i \frac{\partial}{\partial x^i}$. Either way you get $\sum_j \sum_i \frac{\partial^2}{\partial x^i \partial x^j} v^i w^j$. I feel like this only works if I had $dx^i \otimes dx^j$ in my computation. Can someone help clear this up?
 A: I think you're getting understandably confused by the notation for the coordinates. Let's use $x^i$ for the coordinates on W and $v^j$ for the vertical coordinates on $T^\ast W$ (you can think of $v^j$ as the coefficient on $dx^j$, but let's not write $dx^j$ yet). We're identifying $T^\ast U$ with $U \times \mathbb{R}^n$. The vertical part (the $\mathbb{R}^n$ part) of $df$ is expressed by $v^j = \frac{df}{dx^j}$ (and the horizontal part is the identity map). 
Now, $d_p(df): T_p W \to T_{(p,0)}(T^\ast W) \cong T_p W \oplus T_p^\ast W$ is a linear map between vector spaces. Let's make it clear that we're using the basis $\{ \dots, \frac{\partial}{\partial x^i}, \dots \}$ for $T_p W$ and the basis $\{\dots, dx^j, \dots \}$ for $T_p^\ast W$. $d_p(df)$ maps the basis vector $\frac{\partial}{\partial x^i}$ to $\frac{\partial}{\partial x^i} \oplus \sum_j \frac{\partial v^j}{\partial x^i} dx^j$. After projecting onto the second factor, we obtain the linear map $T_p W \to T_p^\ast W$ that maps $\frac{\partial}{\partial x^i}$ to $\sum_j \frac{\partial v^j}{\partial x^i} dx^j = \sum_j \frac{\partial^2 f}{\partial x^i \partial x^j} dx^j$. If you like, you can view this as an element of $T_p^\ast W \otimes T_p^\ast W$ and write it as $\frac{\partial^2 f}{\partial x^i \partial x^j} dx^i \otimes dx^j$, as you suggested.
Note that under the identification, $T_{(p,0)}(T^\ast W) \cong T_p W \oplus T_p^\ast W$, we are identifying the vertical tangent vector $\frac{\partial}{\partial v_j}$ with $dx^j$. Also, I am writing $T_p W$ instead of $T_{(p,0)} Z$ since we can identify the zero section $Z$ with $W$.
By the way, if I'm understanding this correctly, I think the notation should be $(d_p(df))(v)$, not $d_p(df(v))$.
