# Inner product between tangent vectors

Suppose we are given a point $p$ on a Riemannian manifold $M$ with a tangent space $V=T_{p}M$. Furthur assume that this tangent space is Euclidean. Suppose we have two tangent vectors represented in differential geometry as vectors $\mathbf{A}=A^{i}\partial_{i}$and $\mathbf{B}=B^{j}\partial_{j}.$ My question is what would the dot product between two such vectors ($V\times V\rightarrow\mathbb{R})$ look like? In an orthogonal coordinate system, how could we show that $<\partial_{i},\partial_{j}>=\delta_{ij}$?

Note that I am intentionally avoiding one-forms or covectors here, since in linear algebra the scalar product is defined on $V\times V\rightarrow\mathbb{R}$. I know how to form a scalar product between a vector and covector so that's not at issue.

• What is your definition of a Riemannian manifold? As I know it, the looks of $\langle A,B\rangle$ are defined by the implied Riemannian inner product. – M. Luethi Feb 3 '14 at 13:37
• You are right that I have implied a Riemannian inner product. I'm just trying to come to grips with how vectors whose basis are partial derivative operators can give a real number as a scalar product. – user125226 Feb 3 '14 at 13:48
• Then you can use linear algebra: chose a basis in which the inner product is represented by a diagonal matrix (Sylvester's law). Then the coordinate basis becomes a linear combination of the chosen basis. In particular we do not need $\langle\partial_{i},\partial_{j}\rangle$. As an example look up the upper real plane (hyperbolic plain), where in the natural coordinate system $\langle\partial_{y},\partial_{y}\rangle=y^{-2}$. Or you can just choose $\mathbb{R}^{2}$ with something else than the identity for the product. I think $\left(\begin{matrix} 1 & 0.5\\ 0.5 & 1\end{matrix}\right)$ works – M. Luethi Feb 3 '14 at 13:54
• What I would like to understand is how to use these partial derivative operators to obtain the elements of the Riemannian metric. What are these operators operating on? – user125226 Feb 3 '14 at 14:13
• You can use the definition of a Riemannian metric: for every coordinate system the functions $\langle\partial_{i}|_{p}.\partial_{j}|_{p}\rangle_{p}$ are smooth functions. The values of those functions determine at the inner product at $p$ completely because the partial derivatives form a basis of $V$. Now if you want to know the expression for another basis, e.g. $\partial_{1}+\partial_{2},\partial_{2},\ldots,\partial_{n}$, you necessarily need to know the base change matrix. The partial derivatives operate on germs of functions. There is no need to drop the initial understanding of $T_{p}M$. – M. Luethi Feb 3 '14 at 14:20

The tangent space $V = T_pM$ has been endowed with a metric, a bilinear form $g$. Given a basis $\{v_i\}$ for $V$, the metric has coordinates $g_{ij} = g(v_i,v_j)$. In general, the dot product between two vectors $A$ and $B$ is given by $g(A,B)$; in coordinates, with $A = \sum A^iv_i$ and $B = \sum B^jv_j$, then the dot product is $$g(A,B) = \sum_{i,j} g_{ij}A^iB^j.$$ (You can prove this quickly by appealing to bilinearity.)

If you choose an orthonormal basis for $T_pM$, then by definition of orthonormality the coordinates of $g$ are $g_{ij} = \delta_{ij}$. An orthonormal basis always exists, and can be constructed via the Gram-Schmidt process.

Beware: In general an orthonormal basis of $V$ does not come from a local coordinate system on the manifold $M$. That is, for coordinate vector fields, you should not assume that $g(\partial_i,\partial_j)=\delta_{ij}$. The obstruction for this happening in a neighborhood is the curvature tensor.

To elaborate,

• Coordinate functions $x^i:U\to\mathbb{R}$ induce a basis $\frac{\partial}{\partial x^i}$ at each point in the coordinate patch. These induce coordinate functions $g_{ij}$ for the metric. Unless curvature vanishes, $g_{ij}$ will not be equal to $\delta_{ij}$ in any open subset of the patch.
• It is possible to choose a set of vector fields $\{e_i\}$ on an open set of the manifold such that at each point the $\{e_i\}$ form an orthonormal basis of the tangent space. In these bases, the metric will at each point of this open set be $\delta_{ij}$. However, unless curvature vanishes, the orthonormal framing will not be integrable to coordinates. (As above, an orthonormal frame field can be created by applying Gram-Schmidt pointwise to coordinate vector fields.)
• It is possible to choose coordinates so that at a single point the coordinate functions of the metric are $\delta_{ij}$. One way of doing this is normal coordinates. However, unless curvature vanishes, away from that point, in those coordinates the metric will no longer be the identity matrix.

An inner product is a choice of bilinear form $g:V\times V\to \mathbb{R}$. That is, once an inner product has been fixed, you already know how to multiply two tangent vectors. What I explain above is how to find the expression for this multiplication in coordinates on $V$.

Once an inner product has been chosen, then you can define the musical isomorphism via $$v^\flat(w) = g(v,w).$$ So in the sense that the musical isomorphism is defined exactly so that $v^\flat(w) = g(v,w)$, finding $g(v,w)$ involves finding $v^\flat$ and then contracting with $w$, but "morally" the definition of $g$ precedes the definition of $\cdot^\flat$.

For further reading, I recommend O'Neill, Semi-Riemannian Geometry with an Introduction to Relativity. Chapters 1 and 2 cover the basic smooth manifold construction and multilinear algebra that you're asking about here. This may be especially suited to you because I'm guessing you come from a physics background.

Other introductory books to differential geometry will also cover this material at varying levels of abstraction. Do Carmo's book Riemannian Geometry is a standard intro graduate text. Warner's book Introduction to Differentiable Manifolds covers the same material in the first couple of chapters, but more abstractly.

• Aren't you really just lowing an index and converting one of the vectors into a covector (via the musical isomorphism) and performing a contraction? Because this method I understand. – user125226 Feb 3 '14 at 13:52
• @user125226 That's another way of putting it, but the musical isomorphisms are only well-defined because we have already chosen an inner product. – Neal Feb 3 '14 at 14:10
• @user125226 I have added a note about the musical isomorphism. Let me know if you have more questions or if it's not clear. – Neal Feb 3 '14 at 14:47
• Thanks. That makes it clearer. I do have a question about your last paragraph before the addendum ... Does this mean that the partial derivative operators operate on coordinate functions on the manifold and that the basis vectors for the tangent space is different from the basis of the coordinate patch? This is where it gets murky for me. – user125226 Feb 3 '14 at 15:00
• Thanks, that's a big help and much appreciated! You are right, I come from a physics background and want to take a more "modern" approach to vectors and tensors. – user125226 Feb 3 '14 at 19:24