# Musical isomorphism

In mathematics, the musical isomorphism (or canonical isomorphism) is an isomorphism between the tangent bundle TM and the cotangent bundle T*M of a Riemannian manifold given by its metric. There are similar isomorphisms on symplectic manifolds.

It is also known as raising and lowering indices.

## Discussion

Let $(M,g)$ be a Riemannian manifold. Suppose $\{\partial_i\}$ is a local frame for the tangent bundle $TM$ with dual coframe $\{dx^i\}$. Then, locally, we may express the Riemannian metric (which is a 2-covariant tensor field which is symmetric and positive-definite) as $g=g_{ij}\,dx^i \otimes dx^j$ (where we employ the Einstein summation convention). Given a vector field $X=X^i \partial_i$ we define its flat by

$X^\flat := g_{ij} X^i \, dx^j=:X_j \, dx^j.$

This is referred to as 'lowering an index'. Using the traditional diamond bracket notation for inner product defined by g, we obtain the somewhat more transparent relation

$X^\flat (Y) = \langle X, Y \rangle$

for all vectors X and Y.

Alternatively, given a covector field $\omega=\omega_i \, dx^i$ we define its sharp by

$\omega^\sharp :=g^{ij} \omega_i \partial_j$

where $g^{ij}$ are the elements of the inverse matrix to $g_{ij}$. Taking the sharp of a covector field is referred to as 'raising an index'.

Through this construction we have two inverse isomorphisms $\flat:TM \to T^*M$ and $\sharp:T^*M \to TM$. These are isomorphisms of vector bundles and hence we have, for each $p \in M$, inverse vector space isomorphisms between $T_pM$ and $T^*_pM$.

The musical isomorphisms may also be extended to the bundles $\bigotimes ^k TM$ and $\bigotimes ^k T^*M$. It must be stated which index is to be raised or lowered. For instance, consider the (2,0) tensor field $X=X_{ij} \, dx^i \otimes dx^j$. Raising the second index, we get the (1,1) tensor field $X^\sharp = g^{jk}X_{ij} \, dx^i \otimes \partial _k.$

## Trace of a tensor through a metric

Given a (2,0) tensor field $X=X_{ij} \, dx^i \otimes dx^j$ we define the trace of $X$ through the metric $g$ by

$\operatorname{tr}_g(X):=\operatorname{tr}(X^\sharp)=\operatorname{tr}(g^{jk}X_{ij}) = g^{ji}X_{ij} = g^{ij}X_{ij}.$

Observe that the definition of trace is independent of the choice of index we raise since the metric tensor is symmetric.