# I What is difference between the "1. kind" Cristoffel symbol and the "2. kind" Cristoffel symbol?

#### olgerm

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What is difference between 1. kind Cristoffel symbol and 2. kind Cristoffel symbol?
Is the Cristoffel symbol in ricci curvature tensor 1. kind or 2. kind?

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#### olgerm

Gold Member
Is the Cristoffel symbol in ricci curvature tensor 1. kind or 2. kind?
Where should I use 1. kind Cristoffel symbol and where 2. kind Cristoffel symbol?
Wikipedia only gives definitions of 2 types of Cristoffel symbols.

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#### strangerep

Wikipedia only gives definitions of 2 types of Cristoffel symbols.
Yes, but did you actually read those definitions and think about them?? Each kind Christoffel symbol (CS) can be obtained from the other by either raising or lowering the 1st index using the metric.

And did you actually re-look at the Ricci tensor after looking at both kinds of CS?

Of course, the Ricci tensor can be written in terms of either kind of CS. But it's a bit less messy if you use the CS 2nd kind, which is what's usually done in textbooks.So the answer to your "when should I use...?" question is basically "use whatever CS kind is more convenient for whatever you're doing".

(I get the feeling you're really overthinking this, but I can't yet see exactly where your mental block lies.)

BTW, since you seem to insist on mis-spelling it, the correct spelling is "Christoffel".

#### olgerm

Gold Member
Each kind Christoffel symbol (CS) can be obtained from the other by either raising or lowering the 1st index using the metric.
So if first index is up it is 1. kind Christoffel symbol and is first index is down it is 2. kind Christoffel symbol. These are not both same quantities coavariant and contravariant parts beacause Christoffel symbol is not tensor?

#### olgerm

Gold Member
If the difference between "1. kind" Cristoffel symbol and the "2. kind" Cristoffel symbol is just whether 1. index is contravariant or covariant, then why does the same symbol have different names(1. kind - and 2. kind)? tensors are not called 1. or 2. kind depending whther their 1. index is contravariant or covariant.

#### strangerep

So if first index is up it is 1. kind Christoffel symbol and is first index is down it is 2. kind Christoffel symbol. These are not both same quantities coavariant and contravariant parts beacause Christoffel symbol is not tensor?
The phrase "covariant and contravariant parts" has no meaning.

Any quantity, (regardless of whether tensor or not), is not the "same" when you raise or lower one of its indices with the metric.

E.g., suppose we're working with an ordinary vector space $V$. Components of an element $v \in V$ are typically written as $v^\mu$. If we now work with a tensor product space $V\otimes V$, we write the components of an element $h \in V\otimes V$ as $h^{\mu\nu}$.

If we are working with the dual space of $V$, denoted $V^*$, components of an element $u\in V^*$ are usually denoted as $u_\mu$.

For finite-dimensional vector spaces, the primal and dual spaces $V$ and $V^*$ are isomorphic -- we can find a 1-to-1 mapping between them. In relativity, this is usually done with the help of a metric tensor $g \in V^* \otimes V^*$, whose components are denoted $g_{\mu\nu}$ and its inverse $g^{\mu\nu} \in V \otimes V$ (defined via $g_{\mu\alpha} g^{\alpha\nu} = \delta^\nu_\mu$).

Hence our $v^\mu$ (components of an element of $V$) can be mapped to (components of) an element of $V^*$, denoted $v_\alpha$, using the recipe $v_\alpha = g_{\alpha\beta} v^\beta$.

Once again, for emphasis, elements of $V$ and $V^*$ are not a priori the same in any sense. Rather, we impose a 1-to-1 mapping between them using the metric.

The same ideas apply to any object with a mixture of upper and lower indices. E.g., something like $\delta^\nu_\mu$ denotes the components of an element of $V \otimes V^*$.

BTW, if all this stuff about vector spaces, dual spaces, and tensor products, etc, seems like gobbledegook, then you really need to study a suitable textbook. Since you're obviously studying GR, which textbook(s) are you working from?

#### strangerep

then why does the same symbol have different names(1. kind - and 2. kind)?
That's just an historical anomaly. I don't like it, since it tends to confuse people, as you're discovering. I wish the "1st-kind/2nd-kind" nomenclature of Christoffel symbols would just fade away as an historical relic, no longer relevant.

#### robphy

Homework Helper
Gold Member
"... of the first kind", "... of the second kind", etc.. are not particular to Christoffel symbols.

You've got Bessel functions (Hankel functions are sometimes called "Bessel functions of the third kind"),
Elliptic Functions, the Christoffel Symbols, Stirling Numbers, ...
...oh, and don't forget Close Encounters.

#### olgerm

Gold Member
"... of the first kind", "... of the second kind", etc.. are not particular to Christoffel symbols.

You've got Bessel functions (Hankel functions are sometimes called "Bessel functions of the third kind"),
Elliptic Functions, the Christoffel Symbols, Stirling Numbers, ...
...oh, and don't forget Close Encounters.
I think these usagegse of "... of the first kind", "... of the second kind" have different meaning. not about upper or lower index.

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#### pervect

Staff Emeritus
My take on the matter is that the Christoffel sybmols are not tensors. If they were tensors, we'd regard raising or lowering the index as something that didn't matter, the upper index and lower index versions would be regarded as the "same" tensor, just with a different index position. This viewpoint, that tensors are "the same" regardless of index position, relies on having a metric tensor. Without a metric tensor to raise and lower indices, we'd have to think of raising and lowering indices differently.

But Christoffel symbols are not tensors so the above doesn't apply.

I think the two different kinds of Christoffel symbols do represent the "same" connection, in the above sense. But I could be confused about this.

#### vanhees71

Gold Member
No, Christoffel symbols are not tensors nor tensor components but components of a connection.

Tensors have no indices and are invariant objects, i.e., they are independent of the choice of a basis and its dual basis. There are tensor components with upper and lower indices, the one tranforming contravariantly (upper indices) the other transforming covariantly. The corresponding basis vectors have lower indices transforming covariantly, and the associated dual vectors have upper indices transforming contravariantly under changes of bases and co-bases.

#### pervect

Staff Emeritus
Tensors have no indices and are invariant objects, i.e., they are independent of the choice of a basis and its dual basis.
While that's how I think of them, I only work with theories such as GR that have a metric. I'm not sure what the correct approach is if one doesn't have a metric, and thus lacks a natural map between the basis vectors and the dual basis vectors.

For an example of a non-metric theory of gravity, Newton-Cartan theory comes to mind.

#### olgerm

Gold Member
I other tensor properties, like that I can raise and lower same indices on both sides of equation, also true about christoffel symbol?

#### haushofer

While that's how I think of them, I only work with theories such as GR that have a metric. I'm not sure what the correct approach is if one doesn't have a metric, and thus lacks a natural map between the basis vectors and the dual basis vectors.

For an example of a non-metric theory of gravity, Newton-Cartan theory comes to mind.
Newton-Cartan theory is a multi-metric theory, actually :P

#### vanhees71

Gold Member
While that's how I think of them, I only work with theories such as GR that have a metric. I'm not sure what the correct approach is if one doesn't have a metric, and thus lacks a natural map between the basis vectors and the dual basis vectors.

For an example of a non-metric theory of gravity, Newton-Cartan theory comes to mind.
For tensors and vectors you don't need a metric (GR has has a fundamental form but not a metric, but that's another story). They are in any case by definition invariant objects, i.e., independent of any choice of basis.

#### robphy

Homework Helper
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I other tensor properties, like that I can raise and lower same indices on both sides of equation, also true about christoffel symbol?
Yes.
In another recent thread on raising and lowering indices,
robphy said:
First realize that "raising" and "lowering" of indices are really just shortcuts:
Given $v^a$, then $v_b \equiv v^a g_{ab}$.
So, the real object is $v^a g_{ab}$, but it gets tiring to write that all of the time.
So, a convention is adopted to define the shortcut $v_b$.
So, the same holds true for any free index in a tensor expression.

While that's how I think of them, I only work with theories such as GR that have a metric. I'm not sure what the correct approach is if one doesn't have a metric, and thus lacks a natural map between the basis vectors and the dual basis vectors.

For an example of a non-metric theory of gravity, Newton-Cartan theory comes to mind.
Newton-Cartan theory is a multi-metric theory, actually :P
Some strict interpretations of "metric" are restricted only to Riemannian-signature.
A less strict interpretation allows pseudo-Riemannian (Lorentzian signaure, for example).
But one has to be willing to relax things further to allow degenerate [i.e. non-invertible] metrics, like the two degenerate ones in Newton-Cartan.

(From the point of view of Cayley-Klein projective geometry, Euclidean geometry is degenerate, represented by two matrices with signature $(+, 0, 0)$ and $(+, +, 0)$. See pg 392 of https://www-m10.ma.tum.de/foswiki/pub/Lehre/WS0910/ProjektiveGeometrieWS0910/GeomBook.pdf ... type V. Galilean geometry (a flat Newton-Cartan) is type VII. I think 1+1 Newton-Cartan is Type III and type VI depending on the sign of the curvature.)
So, degeneracy isn't all that bad.)

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#### PeterDonis

Mentor
GR has has a fundamental form but not a metric
Do you mean Newton-Cartan theory here? GR has a metric.

#### vanhees71

Gold Member
No GR has not a metric. What physicists call metric is a fundamental form since it's not positive definite but of signature (1,3) or equivalently (3,1). This is a very important point since it allows to describe (local) causal laws.

#### olgerm

Gold Member
First realize that "raising" and "lowering" of indices are really just shortcuts:
Given $v^a$, then $v_b≡v^a*g_{ab}$.
So, the real object is $v^a*g_{ab}$, but it gets tiring to write that all of the time.
So, a convention is adopted to define the shortcut $v_b$.
To get rid of covariance even in metric use relation $g_{ab}=((g)^{-1})^{ab}=(g)^{-1}(ab)$ that covaraint metric is inverse matrix of (contravariant) metric.
$^{-1}$ means here inverse matrix.

So $v_b≡\sum_{a=0}^D(v(a)*g^{-1}(a,b))$

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#### vanhees71

Gold Member
Again you have a very unusual and confusing notation. The co- and contravariant metric components are related by (Einstein summation convention assumed)
$$g_{ab} g^{bc}=\delta_a^c,$$
i.e., the matrix $(g_{ab})$ is the inverse of the matrix $(g^{ab})$.

As any tensor the metric is invariant and doesn't depend on the basis chosen, i.e.,
$$\boldsymbol{g}=g_{ab} \boldsymbol{b}^{a} \otimes \boldsymbol{b}^{b} = g^{ab} \boldsymbol{b}_a \otimes \boldsymbol{b}_b,$$
where $\boldsymbol{b}_a$ are basis vectors and $\boldsymbol{b}^a$ the corresponding dual-basis vectors.

#### olgerm

Gold Member
Again you have a very unusual and confusing notation.
robphy point was that it is possible and good to get rid of covariant component by replacing $v_b$ with $v^a*g_{ab}$.
My point is that now that you have only contravariant components you can replace $v^a$ with $v(a)$ so $\sum_{a=0}^D(v(a)*g^{-1}(a,b))$.

In this notation upper index can be used to notate exponentation.

#### PeterDonis

Mentor
GR has not a metric. What physicists call metric is a fundamental form
Ah, ok, you were using "metric" in the strict positive definite sense, not the sloppy physicist's sense.

#### strangerep

robphy point was that it is possible and good to get rid of covariant component by replacing $v_b$ with $v^a*g_{ab}$.
I should probably let @robphy speak for himself, but I'm reasonably sure you've misinterpreted what he wrote, i.e.,
robphy said:
Given $v^a$, then $v_b \equiv v^a * g_{ab}$.
So, the real object is $v^a∗g_{ab}$, but it gets tiring to write that all of the time.
(My emboldening.)

My point is that now that you have only contravariant components you can replace $v^a$ with $v(a)$ so $\sum_{a=0}^D(v(a)*g^{-1}(a,b))$.
That would just make the mess a lot more tedious.

Anyway, you seem to be drifting away from your original topic of this thread, so maybe it's time to close this thread and start a new one if you really want to pursue this sub-issue (futile though that would be).