# Exponential of the metric tensor

Exponential of an arbitrary matrix can be written as $$e^A = \displaystyle\sum_{n=0}^\infty \dfrac{A^n}{n!}$$

1. In Einstein notation, how this expression will look like?

2. In Einstein notation, what will be the exponential of an arbitrary metric tensor?

My attempt: I guess, (because I'm not find it anywhere), the answer of Qs. $$1$$ will be $$\left(e^A\right)_{ij} = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} A_i^{~k_1} A_{k_1}^{~k_2} ... A_{k_{n-1}j}$$

This answer in Mathematics SE (although unvoted) is also suggesting this expression.

Now if $$A$$ is the metric tensor itself, then this component form of exponential will reduce as

$$\left(e^g\right)_{ij} = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} g_i^{~k_1} g_{k_1}^{~k_2} ... g_{k_{n-1}j}$$

$$\Longrightarrow \left(e^g\right)_{ij} = \displaystyle \sum_{n=0}^\infty \frac{g_{ij}}{n!} = g_{ij} \displaystyle\sum_{n=0}^\infty \dfrac{1}{n!} = e~g_{ij}$$

because metric tensor of mixed indices is essentially the Kronecker delta.

But this shouldn't happen, and can easily be checked by taking the exponential of a diagonal metric (like Schwarzschild metric) !! So am I mistaking in assuming the Einstein notation version of matrix exponentiation? What will be the correct one?

Note that when raising an index of the metric tensor, you get a Kronecker delta (by design): \begin{align} {g_i}^j=g_{ik}g^{kj}=\delta_i^j \end{align} This is because $$g^{kj}$$ is shorthand for the inverse of the metric tensor.

To properly define the matrix exponential, you would have to define the square of the metric tensor. This will be be hard to do, because the metric is different from a "regular" matrix. Let $$V$$ be a vector space and $$V^*$$ be the corresponding covector space. A "regular" matrix (a linear map) sends vectors to vectors: $${M^i}_j:V\mapsto V$$. A metric tensor sends two vectors to a scalar: $$g_{ij}:(V,V)\mapsto \mathbb R$$. The inverse metric tensor $$g^{ij}$$ sends two covectors to a scalar: $$g^{ij}:(V^*,V^*)\mapsto\mathbb R$$.

You can do the regular matrix twice easily, because it sends vectors to vectors. For the metric tensor this becomes awkward.

Bonus question: what do you get when you exponentiate the identity matrix?

• Good point @AccedentalTaylorExpansion.
– SCh
Commented Apr 9 at 15:08
• Exponential of an identity matrix should be Euler's number $(e)$ times identity matrix itself.
– SCh
Commented Apr 9 at 15:10
• Correct! : ) ‎ ‎ ‎ Commented Apr 9 at 15:13
• So what will be the expression for $\exp(g)_{ij}$?
– SCh
Commented Apr 9 at 18:51
• @SCh there is no definition for the exponential of tensors in general. You can only define exponentials of matrices (or more generally, linear operators on a fixed (Banach) space) because you can define a product of matrices (resp. composition of endomorphisms). Hence, there is no canonical meaning of $g^2=g\cdot g$ which produces again a $(0,2)$-tensor; so likewise, there is no $g^3,g^4$ etc, so you can’t define $\exp(g)$. If you really want to, you’re going to need to do something adhoc and define a bilinear map $B:T^0_2(V)\times T^0_2(V)\to T^0_2(V)$, and define products relative to this. Commented Apr 10 at 4:16