Is the Levi-Civita connection of any sum of 2 metrics, the sum of their respective Levi-Civita connections?

[MathNugget]

If g and g' are 2 metrics on the smooth manifold M, $\nabla$ is the Levi-Civita connection of g, $\nabla'$ is the Levi-Civita connection of g'.

Is $\nabla+\nabla'$ the Levi Civita connection of g+g'?

If it was so, then the following 2 properties should work:
a) $(\nabla+\nabla')(g+g')=0$
b) $(\nabla+\nabla')_XY-(\nabla+\nabla')_YX=[X, Y]$

I suspect this isn't true, but then I'd have to find a counterexample. I can't find how these things work, besides some really abstract articles, like the wikipedia one. I'll go out on a limb:

$(\nabla+\nabla')(g+g')=\nabla g + \nabla g' +\nabla' g +\nabla' g' = \nabla' g + \nabla g'$.

Side question: what even is $\nabla g$? I read that the connection applies to vector fields, and the metric applied to vector fields gives a scalar. What is the connection doing to the metric? All I can find is intuitive concepts, like "the Levi Civita connection preserves the metric", and "an affine connection connects tangent spaces". In other words...what would $\nabla g$ even be, if it were not 0? Would it be a number, some constant expression, a vector, a tensor...

 

[fresh_42]

How did you find this question? The metric $g$ measures the distance between two tangent vectors. Your question somehow sounds to me as if you want to measure this distance with two different rulers and then add the results.

Side question: what even is $\nabla g$?

Good question. The Levi-Civita connection applies to vector fields, not to metrics.

 

[MathNugget]

Good question. The Levi-Civita connection applies to vector fields, not to metrics.

Wel, even wikipedia: https://en.wikipedia.org/wiki/Levi-Civita_connection uses the notation $\nabla g=0$. I suppose it is a thing of some sorts...

If g and g' are 2 metrics on the smooth manifold M, $\nabla$ is the Levi-Civita connection of g, $\nabla'$ is the Levi-Civita connection of g'.

Is $\nabla+\nabla'$ the Levi Civita connection of g+g'?

This above was the question I had on an exam. Similarly, there was also one with $-\frac{1}{2}g+\frac{3}{2}g'$ and respectively $-\frac{1}{2}\nabla+\frac{3}{2}g'\nabla'$.

I managed to find on some forum that a linear combination of connections with coefficient sum 1 is a connection (not sure why, but even then: would the sum of 2 L-C be an L-C?)

 

[fresh_42]

Yes, but they do not say what is meant. The German version directly demands
$$X(g(Y,Z))=g(\nabla_Z X,Y) +g(X,\nabla_Z Y)$$without dubious $\nabla g.$ The metric is a function $g\, : \,M \longrightarrow \mathbb{R}.$ with $g(p)\, : \,T_pM\times T_pM \longrightarrow \mathbb{R}$.

The requirement is, that $g$ is smooth, hence $Dg$ exists and defines a vector field on $M.$ The proof on the English Wikipedia uses $\nabla_X (g) =0.$ My suspicion is, that $\nabla g =0$ is a sloppy notation for $\nabla_X(Dg) =0$ for every vector field $X.$

The SE page does not add metrics, only connections that operate on the same vector fields. With my interpretation, we get
$$\nabla_X(D(g+g')) = \nabla_X(Dg) +\nabla_X(Dg')$$The first term on the right is zero since $\nabla_X$ is a Levi-Civita connection for $g.$ However, we have no information what the Levi-Civita connection defined by the metric $g$ does on a different metric, and vice versa for $(\nabla', g').$ My answer would be a no, because we don't get a grip on the mixed terms.

 

[MathNugget]

Yes, but they do not say what is meant. The German version directly demands
$$X(g(Y,Z))=g(\nabla_Z X,Y) +g(X,\nabla_Z Y)$$without dubious $\nabla g.$ The metric is a function $g\, : \,M \longrightarrow \mathbb{R}.$ with $g(p)\, : \,T_pM\times T_pM \longrightarrow \mathbb{R}$.

The requirement is, that $g$ is smooth, hence $Dg$ exists and defines a vector field on $M.$ The proof on the English Wikipedia uses $\nabla_X (g) =0.$ My suspicion is, that $\nabla g =0$ is a sloppy notation for $\nabla_X(Dg) =0$ for every vector field $X.$

The SE page does not add metrics, only connections that operate on the same vector fields. With my interpretation, we get
$$\nabla_X(D(g+g')) = \nabla_X(Dg) +\nabla_X(Dg')$$The first term on the right is zero since $\nabla_X$ is a Levi-Civita connection for $g.$ However, we have no information what the Levi-Civita connection defined by the metric $g$ does on a different metric, and vice versa for $(\nabla', g').$ My answer would be a no, because we don't get a grip on the mixed terms.

I think you're right. I was trying to build a counterexample, and I think it might be easier to show in coordinates, with the Christoffel symbols.
Now, I have to build g and g'. Here's my idea:

We pick $\mathbb{R}^3$ to be the manifold.

I know g=$\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$ is a riemannian metric. (can't prove it though, but let's say we know it)

With some wikipedia study, I was able to create this matrix, that I claim to be positive-definite and simmetric:
g'=$\begin{pmatrix} 5 & 4 & 0\\ 4 & 5 & 0\\ 0 & 0 & 4 \end{pmatrix}$ with the inverse $\begin{pmatrix}\frac{5}{9} & -\frac{4}{9} & 0\\ -\frac{4}{9} & \frac{5}{9} & 0\\0 & 0 & \frac{1}{4}\end{pmatrix}$.

I don't exactly know if it's enough for that g' to be a metric.

And now, the Christoffel symbols: $\Gamma^i_{kl}=\frac{1}{2}g^{im}(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{ml}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m})$

What's between the brackets works with (g+g'), but I suspect the inverse matrix $g^{im}$ doesn't work like that.

I think if I prove $(g+g')^{-1}\neq g^{-1}+g'^{-1}$ I show that the Christoffel symbols are different, so $\nabla + \nabla'$ is not the L-C connection. I did the calculations on those matrixes, and this seems to be true.

 

[fresh_42]

I'm not sure whether the abelian Lie multiplication on $\mathbb{R}^3$ is already sufficient.
$$\begin{array}{lll} (\nabla_X + \nabla'_X)(Dg+Dg')&=\nabla_X Dg' +\nabla'_X Dg\\[12pt] &=\nabla_{Dg'}X +[X,Dg'] +\nabla'_{Dg}X +[X,Dg]\\[12pt] &=(\nabla_{Dg'}+\nabla'_{Dg})X+[X,Dg'+Dg] \\[12pt] &=(\nabla_{Dg'}+\nabla'_{Dg})X+[X,D(g'+g)] \end{array}$$
In case of an abelian Lie algebra as in your example of $\mathbb{R}^3 ,$ we get
$$(\nabla_X + \nabla'_X)(Dg+Dg')= (\nabla_{Dg'}+\nabla'_{Dg})X$$ and only have to choose $g'$ such that $\nabla_{Dg'}+\nabla'_{Dg}\neq 0.$

The smallest example of a non-abelian Lie algebra is $\mathfrak{g}=\bigl\langle X,Y\,|\,[X,Y]=0 \bigr\rangle$ which is the tangent space of the Lie group

$$G=\Bigl\langle \begin{pmatrix} e^t&0\\0&e^{-t} \end{pmatrix}\, , \,\begin{pmatrix} 1&c\\0&1 \end{pmatrix}\,|\,t,c\in \mathbb{R} \Bigr\rangle$$
This is clearly a Riemannian manifold that has two major advantages. It has only two dimensions, and the Lie algebra has no center. The latter makes sure that $[X,D(g+g')]\neq 0$ since otherwise $D(g+g')=0$ which we can easily exclude by the choice of $g'.$

This gives you two possible counterexamples: either $\nabla_{Dg'}+\nabla'_{Dg}\neq 0$ for abelian Lie algebras, or $D(g+g')\neq 0$ for non-abelian Lie algebras. Of course, there is still the unlikely case that both terms cancel each other, but that would be really unlucky.

 

[MathNugget]

I'm not sure whether the abelian Lie multiplication on $\mathbb{R}^3$ is already sufficient.
$$\begin{array}{lll} (\nabla_X + \nabla'_X)(Dg+Dg')&=\nabla_X Dg' +\nabla'_X Dg\\[12pt] &=\nabla_{Dg'}X +[X,Dg'] +\nabla'_{Dg}X +[X,Dg]\\[12pt] &=(\nabla_{Dg'}+\nabla'_{Dg})X+[X,Dg'+Dg] \\[12pt] &=(\nabla_{Dg'}+\nabla'_{Dg})X+[X,D(g'+g)] \end{array}$$
In case of an abelian Lie algebra as in your example of $\mathbb{R}^3 ,$ we get
$$(\nabla_X + \nabla'_X)(Dg+Dg')= (\nabla_{Dg'}+\nabla'_{Dg})X$$ and only have to choose $g'$ such that $\nabla_{Dg'}+\nabla'_{Dg}\neq 0.$

The smallest example of a non-abelian Lie algebra is $\mathfrak{g}=\bigl\langle X,Y\,|\,[X,Y]=0 \bigr\rangle$ which is the tangent space of the Lie group

$$G=\Bigl\langle \begin{pmatrix} e^t&0\\0&e^{-t} \end{pmatrix}\, , \,\begin{pmatrix} 1&c\\0&1 \end{pmatrix}\,|\,t,c\in \mathbb{R} \Bigr\rangle$$
This is clearly a Riemannian manifold that has two major advantages. It has only two dimensions, and the Lie algebra has no center. The latter makes sure that $[X,D(g+g')]\neq 0$ since otherwise $D(g+g')=0$ which we can easily exclude by the choice of $g'.$

This gives you two possible counterexamples: either $\nabla_{Dg'}+\nabla'_{Dg}\neq 0$ for abelian Lie algebras, or $D(g+g')\neq 0$ for non-abelian Lie algebras. Of course, there is still the unlikely case that both terms cancel each other, but that would be really unlucky.

I have trouble understanding Lie algebras, and matrices as manifolds in general...

I've been doing some research too, and apparently $g^{ij}g_{jk}=\delta^i_k$, so the matrix $g_{ij}$ seems to be 0 for all $i \neq j$ (it did happen to the dot product already, but I thought maybe a riemannian metric doesn't have to in general)...but at the same time, if that is the case, looks like $g^{12}g_{21}=1$, and if $g_{21}=0$, this makes absolutely no sense to me...

How are $g^{ij}$ calculated?

 

[fresh_42]

I have trouble understanding Lie algebras, and matrices as manifolds in general...

You can just differentiate paths on them to get to the tangent spaces.

$$\left. \dfrac{d}{dt}\right|_{t=0} \begin{pmatrix} e^t&0\\0&e^{-t} \end{pmatrix}=\begin{pmatrix} 1&0\\0&-1 \end{pmatrix}\, , \,\left. \dfrac{d}{dt}\right|_{c=0}\begin{pmatrix} 1&c\\0&1 \end{pmatrix}=\begin{pmatrix}0 &1 \\0& 0\end{pmatrix}$$
and you are already in the Lie algebra with

$$\left[\begin{pmatrix} 1&0\\0&-1 \end{pmatrix},\begin{pmatrix} 0&1\\0&0 \end{pmatrix}\right]= 2\cdot\begin{pmatrix} 0&1\\0&0 \end{pmatrix}$$
Of course, the group is larger and there are more paths on it and more points than the identity where I differentiated. But the generators and basis elements are as just described.

I've been doing some research too, and apparently $g^{ij}g_{jk}=\delta^i_k$, so the matrix $g_{ij}$ seems to be 0 for all $i \neq j$ (it did happen to the dot product already, but I thought maybe a riemannian metric doesn't have to in general)...but at the same time, if that is the case, looks like $g^{12}g_{21}=1$, and if $g_{21}=0$, this makes absolutely no sense to me...

How are $g^{ij}$ calculated?

I am really bad with coordinates and don't even know the difference between $g^{ij}$ and $g_{ij}$ without looking it up, e.g. on https://en.wikipedia.org/wiki/Metric_tensor or https://en.wikipedia.org/wiki/Raising_and_lowering_indices. It's a typical physicist's thing. I tried to get used to it but I always fall back to coordinate-free descriptions and already the Einstein notation causes me headaches. But what I do know is, that $g^{ij}g_{jk}=\sum_{j=1}^n g^{ij}g_{jk}= {\delta^i}_k$ so your conclusion about the matrix elements $g_{ik}$ to be zero is not correct. The matrix elements of the product matrix at position $(i,k)$: $(g^{ij} g_{jk} )_{ik} = ( \sum_{j=1}^n g^{ij} g_{jk})_{ik}= 0$ are zero for all $i\neq k.$ You should never use the same index $j$ as a summation index and then as a fixed position. That's confusing and almost certainly causes mistakes sooner or later.

If I were to study it, I would fight my way through both examples, $\mathbb{R}^3$ and $G,$ with both of your metrics, the dot product $g=\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1 \end{pmatrix}$ and $g'=\begin{pmatrix} 5&4&0\\4&5&0\\0&0&4 \end{pmatrix} ,g'=\begin{pmatrix} 5&4\\4&5 \end{pmatrix}$ (hoping that is still positive definite), resp. in the case of $G .$

Maybe I would take the 2-sphere instead of $\mathbb{R}^3.$ There are pictures on the internet about what the Levi-Civita connection does on the 2-sphere. This is one way to study it. I have also seen students who directly learned the index acrobatic disregarding what it all means. Personally, I need examples to learn things. My starting point would be the smooth functions $g,g'\, : \,M \rightarrow \mathbb{R}.$ What is $g'(p)(\partial_i ,\partial_j)$?

 

[MathNugget]

My starting point would be the smooth functions $g,g'\, : \,M \rightarrow \mathbb{R}.$ What is $g'(p)(\partial_i ,\partial_j)$?

Wouldn't that just be $g'_{ij}$?

 

[fresh_42]

Wouldn't that just be $g'_{ij}$?

I'm not sure. If so, then e.g. $g'_{12}=5$ at all points of the manifold. But then, $Dg'=0$ and also $Dg=0.$ With that, we get $(\nabla_X +\nabla'_X)(Dg+Dg')=0$ and $\nabla+\nabla'$ is indeed the Levi-Civita connection of $g+g'$ because everything on the right becomes zero.

However, that the metrics would be independent of the location has to be proven. Consider the example of $G$ again and e.g.
$$p=\begin{pmatrix} e^{t_p}&c_p\\0& e^{-t_p}\end{pmatrix} \in G .$$ What are two (linearly independent) tangent vectors $X_p,Y_p$ at $p$ for which we can compute $g'(p)(X_p,Y_p)$? Everything is fine at the identity matrix, but what happens elsewhere? I'm not sure whether the left-invariance of the vector fields already solves these issues. What we have is that
$$X_p=(DL_p)X_e$$ with $L_p(q)=pq\in G$ and $X_e$ is a tangent vector at $e= \begin{pmatrix} 1&0\\0&1 \end{pmatrix}$ so we need to know the derivative of the left-multiplication $L_p$ to compute $X_p.$ This formula must hold for $X_p$ so it can be used to check whether there were mistakes in case we compute them with paths through $p.$

It is a bit messy and I don't want to say something wrong. That's why I would compute the examples. If both metrics are constant everywhere, then the question becomes trivial and the answer is yes.

By the left-invariance of the vector fields, we get
$$g'(p)(X_p,Y_p)=g'(p)((DL_p)X_e,(DL_p)Y_e)=g'(p)((DL_p)\partial_t,(DL_p)\partial_c)$$but what's next? This is probably when people started to think about connections! How can we compare tangent vectors at two different locations, $e$ and $p$?