# Definition of Kronecker delta

1. Jul 15, 2008

### 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?

2. Jul 16, 2008

### 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...

3. Jul 16, 2008

### Mentz114

I find that,

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

but I can't prove it.

4. Jul 16, 2008

### CompuChip

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

5. Jul 16, 2008

### Mentz114

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

M

6. Jul 16, 2008

### shoehorn

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.

7. Jul 17, 2008

### DrGreg

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.

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