Musical isomorphisms in $\Bbb R^n$ and on a Riemannian manifold

Question

Given a Riemannian manifold $(M,g)$ the musical isomorphisms are defined by $\flat: TM \to T^*M$ with $X^\flat(Y)=g(X,Y)$ and $\sharp: T^*M \to TM$ with $g(\omega^\sharp,Y)=\omega(Y)$.

Contrastly the musical isomorphisms in $\Bbb R^n$ are defined such that if $v = \begin{bmatrix} v_{1} \\ v_{2} \\ \vdots \\ v_{n} \end{bmatrix}$ then $$v^\flat = \begin{bmatrix} v_{1} & v_{2} & \dots & v_{n} \end{bmatrix}$$

and similarly if $w = \begin{bmatrix} w_{1} & w_{2} & \dots & w_{n} \end{bmatrix}$ then $$w^\sharp = \begin{bmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{bmatrix}.$$

I have some trouble seeing how these two are describing the same thing. The definitions in $(M,g)$ are defined using the metric, but in $\Bbb R^n$ this doesn't seem to be the case?

Answer 1

$\newcommand\R{\mathbb{R}}\newcommand\Rh{\hat{\R}}$ The notation I use on a manifold is the following:

Below, let $T = T_M$ denote the tangent space at a point $p$ in a manifold $M$ (which could be $\R^n$. Let $(b_1, \dots, b_n)$ denote any basis of $T$ and $(\beta^1, \dots, \beta^n)$ be the dual basis of $T^*$. All of the components of vectors and covectors below are with respect to these bases.

A tangent vector is represented by a column vector $$ v = \begin{bmatrix} v^1 \\ \vdots \\ v^n \end{bmatrix}. $$ A cotangent vector is represented by a row vector $$ \omega = \begin{bmatrix} \omega_1 & \cdots & \omega_n \end{bmatrix}. $$ The evaluation of a cotangent vector $\omega$ with a tangent vector $v$ is matrix multiplication, \begin{align*} \langle\omega,v\rangle &=\begin{bmatrix} \omega_1 & \cdots & \omega_n\end{bmatrix} \begin{bmatrix} v^1 \\ \vdots \\ v^n \end{bmatrix}\\ &= \omega^iv_i. \end{align*} Everything above works for an arbitrary basis and therefore with respect to any coordinates on a manifold without any Riemannian metric. Duality between tangent and cotangent vectors is represented by column versus row vectors.

Now assume there is a Riemannian metric. This defines an inner product on $T$. Let $(e_1, \dots, e_n)$ be any orthonormal basis of $T$ and $(\omega^1, \dots, \omega^n)$ be the dual basis. All of the components of vectors and covectors below are with respect to these bases. Then $\sharp$ and $\flat$ are simply the tranpose operation. If $v$ is a tangent vector, then $$ v^\flat = v^T $$ and if $\xi$ is a covector then $$ \xi^\sharp = \xi^T. $$ The dot product is simply $$ v\cdot w = v^Tw = v^\flat w $$ and $$ \xi\cdot \eta = \xi \eta^T = \xi\eta^\sharp $$

This can all be generalized pretty easily to formulas with respect to an arbitrary not necessarily orthonormal basis, but that's a longer story.

Answer 2

$v^\flat = \begin{bmatrix} v_{1} & v_{2} & \dots & v_{n} \end{bmatrix} \in (\mathbb{R}^n)^*$ acts on a vector $x \in \mathbb{R}^n$ by $v^\flat (x) := v^\flat \cdot x$.

Here $\cdot$ denotes Matrix multiplication. Of course this is a slight abuse of notation, since we identify the "lying down" vector $v^\flat$ with the "matrix multiplication with $v^\flat$" linear map. By definition $v^\flat \cdot x = \langle v, x\rangle$, where $\langle -, - \rangle$ is the usual scalar product on $\mathbb{R}^n$.

The $\sharp$ case is similar.

Note that we are talking about vector spaces here and not about a Manifold. Otherwise we would need to talk about elements of the tangent (or cotangent) bundle of $\mathbb{R}^n$ and not about elements of $\mathbb{R}^n$ itself.