# A Question about Holonomy of metric connecton

1. Jul 16, 2016

### lichen1983312

if parallel transporting a vector around a loop induces a linear map (an element of holonomy group)
$${P_c}:{T_p}M \to {T_p}M$$

the holonomy group should be a subgroup of
$$GL(m,R)$$

then the book said for a metric connection, the property
$${g_p}({P_c}(X),{P_c}(Y)) = {g_p}(X,Y)$$
makes the holonomy group a subgroup of $$O(m)$$ if the manifold is Riemannian; and a subgroup of $$O(m-1)$$ if the manifold is Lorentzian.

The author must think this is very straightforward and didn't explain why. Can anybody help?

2. Jul 16, 2016

### The Bill

I assume the book you're referring to is Nakahara's Geometry, Topology and Physics, 2e.
This isn't what the book says. Nakahara actually writes ${g_p}({P_c}(X),{P_c}(X)) = {g_p}(X,X)$

In any case, a linear transformation which preserves the inner product is the definition of an orthogonal transformation.

Also, Nakahara doesn't say that the holonomy group is a subgroup of $O(m-1)$ in the Lorentzian case, he says it's a subgroup of $SO(m-1,1)$.

3. Jul 17, 2016

### lavinia

If the connection is compatible with the metric, then parallel translation preserves the metric..

If the vector fields $Y$ and $Z$ are parallel along a curve with tangent vector $X$ then

$X⋅<Y,Z> = <∇_{X}Y,Z> + <Y,∇_{X}Z>$ so the inner product of $Y$ with $Z$ is constant along the curve.

If the metric is Riemannian, that is: it is positive definite on each tangent space, then parallel translation is an orthogonal linear map. If the metric is not positive definite then parallel translation is an element of $O(m-p,p)$. In the case of a Lorentz metric $p=1$.

- If the manifold is not orientable then then a holonomy transformation may be orientation reversing.

4. Jul 18, 2016

### lichen1983312

Thanks for replying, this is very helpful, and I will try to keep the book close next when I am typing.

5. Jul 19, 2016

### lichen1983312

Now I seem to be able to understand the inner product preserving property make a orthogonal group, but can you explain where does the dimension (m-1) come from for Lorenzian case? Thanks

6. Jul 19, 2016

### The Bill

The m-1 comes from the fact that the inner product in the Lorentzian case is usually written with signature (m-1,1). That is, (-1,1,1,1) or (1,-1,-1,-1,-1) for 4d spacetime, for example. In the genral case, this is the indefinite orthogonal group, and Nakahara is only considering the connected component which contains the identity. https://en.wikipedia.org/wiki/Indefinite_orthogonal_group

7. Jul 19, 2016

### lichen1983312

Thanks very much, now I see the point.