Is Kronecker Delta a Coordinate-Independent Identity Operator?

[jdstokes]

Can anyone give me a coordinate-independent definition of $\delta^a_b$ on curved manifolds?

Should it be defined as $\delta^a_b = g^{ac}g_{bc}$ where abstract index notation has been used?

 

[CompuChip]

You could define it that way. Then again, you could take that as the definition for $g^{ab}$ (the inverse metric)
One problem, is that we are used to thinking of the Kronecker delta as "the thing which is 1 iff the indices are equal, and 0 otherwise" which of course, introduces coordinates right away. I am wondering if "the unit tensor" (e.g. dxd unit matrix) is a coordinate independent statement...

 

[Mentz114]

I find that,

$$g_b^a = g^{an}g_{nb} = \delta_b^a$$

but I can't prove it.

 

[CompuChip]

I though that was the definition of the inverse metric. Basically you have written down that
$$g^{-1} g = I$$

 

[Mentz114]

Yep. Going round in cicles. Break the circle at any point and select a definition.

M

 

[shoehorn]

Can anyone give me a coordinate-independent definition of $\delta^a_b$ on curved manifolds?

Should it be defined as $\delta^a_b = g^{ac}g_{bc}$ where abstract index notation has been used?

If your manifold has a metric, you can give a perfectly good coordinate-independent definition of the Kronecker tensor (and, indeed, its generalizations) in terms of the so-called "musical isomorphism" between the tangent space and cotangent space.

This is pretty basic stuff, but beyond a yearning for strict coordinate-independence, I can't see any actual advantage in using such a definition.

 

[DrGreg]

I am wondering if "the unit tensor" (e.g. dxd unit matrix) is a coordinate independent statement...

You are nearly there. The coordinate-free version of a matrix is a linear operator, a function that maps vectors to vectors (or covectors to covectors).

So the Kronecker delta is just the identity operator $\delta (\textbf{X}) = \textbf{X}$ acting on any tangent space (or cotangent space).

This follows from the coordinate expression $\delta^a_b X^b = X^a$ which is true in every coordinate system.