Sometimes am having a problem accessing the thread..what a glitch...

17 hours ago, Markus Hanke said:

What do you mean by this, exactly?

Spacetime is not embedded in anything else, so when we are talking of geometry in this context, what we are referring to is intrinsic geometry on the manifold.

Riemann tensor has 256 components in 4d, Ricci tensor gets the averages/proportionality from Riemann tensor that tend to be more surface topological,Ricci tensor is more critical in constructing Einstein monifold..hope am not wrong on this.

What I meant by interior of the monifold is the other aspects that may have not been captured while getting Ricci  tensor.

7 minutes ago, MJ kihara said:

Riemann tensor has 256 components in 4d

Only 20 components are independent. The Ricci tensor has 10 independent components leaving 10 independent components for the Weyl conformal tensor.

 

Edited October 19, 20241 yr by KJW

On 10/17/2024 at 4:12 PM, MigL said:

Energy is  determined by the configuration of a system ( think potential energy ), so different configurations will have differing energies, even though the systems are comprised of the same individual parts.

Is mass linear or nonlinear i.e don't we add up together mass?...E=mc^2 , individual parts with their masses doesn't their total mass give mass of total of the configuration....the discussion on thread page 5 is spinning off my head ...

40 minutes ago, MJ kihara said:

Riemann tensor has 256 components in 4d

That’s true - but it also has many symmetries in GR, hence only 20 of those are functionally independent.

42 minutes ago, MJ kihara said:

Ricci tensor gets the averages/proportionality from Riemann tensor that tend to be more surface topological,Ricci tensor is more critical in constructing Einstein monifold

The Riemann tensor can be decomposed into two parts - the Ricci tensor, and the Weyl tensor.

Suppose you have a small ball of test particles freely falling in a spacetime. The Ricci tensor tells you how fast the volume of this ball changes with time, whereas the Weyl tensor measures how the shape of the ball gets distorted as it falls.

In vacuum, shapes get distorted, but volumes are preserved during free fall (hence the Ricci tensor vanishes, but the Weyl tensor doesn’t). In the interior of sources, both shapes and volumes may change.

32 minutes ago, MJ kihara said:

Is mass linear or nonlinear i.e don't we add up together mass?

Mass itself adds linearly, but the source term in the field equations is not mass, but the energy-momentum tensor. 

Your looking at Migl's statement wrong.

Let's use an simple everyday world example.

Let's use an electrical circuit.

Take a multimeter on a wire conducting some current. If you take the leads to two point is on the wire itself you cannot measure a voltage.

However if you add some load via say a resistor you can. That is an everyday example of potential difference.

Now let's apply that to our spacetime field. If every coordinate on that field has precisely the same potential energy (potential energy is energy due to location under field treatment) then gravity effectively is zero.

However if there is potential energy differences between coordinate A and coordinate B such as due to a center of mass then you have a gravity term in Newton terms the gradient.

 Now under GR using the full equation \[E^2=(pc^2+m_o c^2)^2\] 

When applied to every coordinate of a field you immediately recognize that both massless particles as well as massive particles can affect the geometry. However the equation also includes their momentum terms so their vectors or spinors also are involved.

Under the stress energy momentum tensor the energy density is the \(T_(00)\) component. The diagonal components, (orthogonal components are the Maximally symmetric components) 

However there are off diagonal components stress, strain,and vorticity these components have symmetry to gas and fluid flow ie through a pipe for everyday examples.

So take a vector field of particles each particle follows its own geodesic those geodesics can converge or diverge from one another as they do so they generate non linearity as they induce curvature terms. When curvature occurs you are naturally inducing acceleration (direction change is also part of acceleration its not just the change in the velocity magnitude) 

Edit cross posted with Markus were both providing the same answer 

(Stress energy momentum tensor relations of a multi particle field).

To make things more complicated each particle of the above field example is interacting with other particles so now our field now has continous changes in velocity.

We can now only average all the numerous curved paths (linearization of a nonlinear system) which is never exact.

The above demonstrates why a tensor field is required. As mentioned a tensor field includes magnitude, vector and spinor relations.

The same applies to a curve you can only average the length of the curve. The extrenums (Maxima and minima) a function  is always a graph (but not all graphs are functions for a graph to have a function it must pass a horizontal and vertical test (off topic).

 

https://tutorial.math.lamar.edu/classes/calcI/minmaxvalues.aspx#:~:text=The function will have an,domain or at relative extrema.

Edited October 19, 20241 yr by Mordred

🙏 Thanks for the answers and explanations above... nothing else I can say more than that.

Your welcome its often tricky to see beyond the mathematics so we're glad to help.

Lol I lost count on how many times getting lost in the math and lose sight of what the math is representing so can readily understand the difficulty 

Edited October 19, 20241 yr by Mordred

On 10/18/2024 at 10:00 AM, MJ kihara said:

Meaning that Ricci tensor may reduce to zero but other aspects of Riemann tensor are still contributing to the manifold.

On 10/19/2024 at 6:11 AM, MJ kihara said:

What I meant by interior of the monifold is the other aspects that may have not been captured while getting Ricci  tensor.

On 10/18/2024 at 12:54 PM, Markus Hanke said:

What do you mean by this, exactly?

Spacetime is not embedded in anything else, so when we are talking of geometry in this context, what we are referring to is intrinsic geometry on the manifold.

On 10/19/2024 at 6:16 AM, KJW said:

Only 20 components are independent. The Ricci tensor has 10 independent components leaving 10 independent components for the Weyl conformal tensor.

 

It has been a while,some clarification from what I asked about the interior of a manifold, Einstein tensor is obtained from ricci tensor which is a contraction of Riemannian tensor....while finding the Riemannian tensor the vector is parallel transported across a parallelogram loop which is assumed to be perfectly closed meaning it's lie bracket is zero ....is this not against special theory of relativity? I.e you can't parallel transports a vector instantly...the lie bracket I think should be approximately zero but not exactly zero ..therefore, Riemann tensor used in GR will leave some information not captured.

2 hours ago, MJ kihara said:

Einstein tensor is obtained from ricci tensor which is a contraction of Riemannian tensor

Yes. Or else you can also calculate it directly from the metric.

2 hours ago, MJ kihara said:

while finding the Riemannian tensor the vector is parallel transported across a parallelogram loop which is assumed to be perfectly closed

This is one way to define the Riemann tensor, though it doesn’t have to be a parallelogram…any closed curved will do.

2 hours ago, MJ kihara said:

is this not against special theory of relativity? I.e you can't parallel transports a vector instantly

Firstly, the space the parallel transport is performed on is already 4-dimensional spacetime, so in general you’re transporting in space and time. Secondly, the crucial point is whether the transported vector, once it returns to its starting point, coincides with its original version before parallel transport, or not. In a curved spacetime, it generally won’t, so the Riemann tensor will allow you to find the difference between the two (ie a new vector that points from the tip of the first vector before parallel transport to the tip of that same vector after parallel transport). All of these vectors are 4-vectors in spacetime.

2 hours ago, MJ kihara said:

Riemann tensor used in GR will leave some information not captured.

In GR, the connection on the manifold is given, being the Levi-Civita connection. Since this connection is torsion free, the only dynamics can come from curvature, so the Riemann tensor captures the entire geometry on your spacetime.

Note though that the Einstein equations in GR do not by themselves completely fix the Riemann tensor (or even the metric). You always need to supply extra boundary conditions to uniquely determine the geometry.

4 hours ago, Markus Hanke said:

Firstly, the space the parallel transport is performed on is already 4-dimensional spacetime, so in general you’re transporting in space and time. Secondly, the crucial point is whether the transported vector, once it returns to its starting point, coincides with its original version before parallel transport, or not. In a curved spacetime, it generally won’t, so the Riemann tensor will allow you to find the difference between the two (ie a new vector that points from the tip of the first vector before parallel transport to the tip of that same vector after parallel transport). All of these vectors are 4-vectors in spacetime.

Am thinking that the loop on which the vector is being parallel transported Is infinitesimal and given the speed of light in a vacuum is huge...the result will appear as the vector has been moved instantly I.e 0.0000000 lots of zero that makes lie bracket taken as zero to give the required answer..what I mean is that the parallel transported vector will end up conciding with the initial vector,therefore,detecting no curvature and since Riemann tensor is the bedrock of GR that effect is carried all the way to Einstein field equations.

Implications of these is that GR is unable to detect zero point energy and quantum fluctuations on the background.

16 hours ago, MJ kihara said:

Implications of these is that GR is unable to detect zero point energy and quantum fluctuations on the background.

All these is compensated by cosmological constant in Einstein field equations,esp if we consider ricci tensor to be zero...the cosmological constant and metric tensor will equate to Einstein constant * energy momentum tensor.

My thinking is that vacuum energy density is related to/is the energy content of graviton,as a side note, "I intent to include all these thinking in the second edition of the book,"the reason for this, is that to prove second Bianchi identity one way is to solve the equation at a point where Christoffel symbols reduces to zero and the metric tensor becomes the Kronecker delta,from my earlier arguments in the thread..where I stated in two dimensions, limit of a curvature is a straight line and a point...having linked that point to a graviton...I can confidently state that gravity is renormalizable by introduction of a graviton,,,,,,and THAT the descripancy between observed cosmological constant and theoretical prediction of zero point energy is due to the rich world inside the graviton that makes divergence to be persistent at zero point.

17 hours ago, MJ kihara said:

what I mean is that the parallel transported vector will end up conciding with the initial vector

No it won’t. On a curved manifold the initial and final vectors can’t coincide. This result is easily shown, and entirely independent of any physics models such as GR. You need to remember that differential (Riemann) geometry was already well established long before Einstein - GR simply uses this discipline of mathematics to formulate its framework.

17 hours ago, MJ kihara said:

the result will appear as the vector has been moved instantly

There is no notion of “duration” associated with this; you simply compare the effects of the connection on your manifold to the original vector.

Another, perhaps simpler, way to state the same thing is that on curved manifolds, covariant derivatives do not commute, which is likewise easily shown.

17 hours ago, MJ kihara said:

Implications of these is that GR is unable to detect zero point energy and quantum fluctuations on the background.

GR is by design a purely classical model, it describes no quantum effects.

1 hour ago, MJ kihara said:

I can confidently state that gravity is renormalizable by introduction of a graviton,

The symmetry group associated with GR (cosmological constant or not) is non-compact and infinite-dimensional; any spin-2 quantum field theory based on such a symmetry group will be non-renormalisable.

7 hours ago, Markus Hanke said:

No it won’t. On a curved manifold the initial and final vectors can’t coincide. This result is easily shown, and entirely independent of any physics models such as GR. You need to remember that (Riemann) geometry was already well established long before Einstein - GR simply uses this discipline of mathematics to formulate its framework.

  On 5/11/2025 at 2:22 PM, MJ kihara said:

When the measurements are at a point in the manifold the Christoffel symbols reduces to zero...i think it depends to the extend and the knowledge we have or we can gain concerning the manifold and it's curvature,for instance, the observable universe is said to be 5%...also to prove if the vector coincide ,it's practically performed,this measurements is limited to the speed of light.its highly possible for the vectors to coincide while there is hidden curvature... information is critical and how we are obtaining it.

Riemann geometry I understand it preceded GR, however,lie bracket is an important part of it and when the loop is completely closed it's taken to be zero,since am looking at away of unifying gravity and quantum mechanics/QFT and given GR is the best framework of how gravity behave I think this is the appropriate point to join the two theories.

9 hours ago, Markus Hanke said:

There is no notion of “duration” associated with this; you simply compare the effects of the connection on your manifold to the original vector.

I think the laws of physics should be universal and if you are parallel transporting a vector,,even if it's in a thought experiment or done mathematically on a sheet of paper we need to consider those laws...like the constancy of speed of light if not so appropriate reasons should be given to explain otherwise.

10 hours ago, Markus Hanke said:

GR is by design a purely classical model, it describes no quantum effects.

Am trying to unify GR with quantum effects ..for instance am stating that divergences at zero point in GR that appear to be non renormalizable are just normal issues of quantum fluctuations that are as a result of rich world inside a graviton....if we equate all this to information, graviton represent the threshold at which this information(hidden curvature) inside the graviton leaks to the outside curvature that becomes detectable by GR i.e a point in the manifold.

10 hours ago, Markus Hanke said:

The symmetry group associated with GR (cosmological constant or not) is non-compact and infinite-dimensional; any spin-2 quantum field theory based on such a symmetry group will be non-renormalisable.

Being infinite dimensional, I think it will depend on the properties of the graviton topology that is also related to quantum fluctuations.

13 hours ago, MJ kihara said:

When the measurements are at a point in the manifold the Christoffel symbols reduces to zero

But this is not what we are doing.

13 hours ago, MJ kihara said:

also to prove if the vector coincide ,it's practically performed,this measurements is limited to the speed of light

And what you’ll find is that they don’t coincide, just as the maths say.

13 hours ago, MJ kihara said:

its highly possible for the vectors to coincide while there is hidden curvature

There’s no such thing as “hidden curvature”. If the manifold is curved, the vectors can’t coincide - that’s all there is to it.

13 hours ago, MJ kihara said:

lie bracket is an important part of it and when the loop is completely closed it's taken to be zero

I don’t know where you are getting this from, but if makes no sense. Given the Levi-Civita connection on a manifold with curvature, the Lie bracket of two vector fields measures the extent to which differential operators associated with these fields fail to commute; specifically in this case, it’s related to the commutator of covariant derivatives, which on a Riemann manifold is not zero, unless the manifold is perfectly flat.

In other words, parallel transport is path-dependent in a curved spacetime.

13 hours ago, MJ kihara said:

Am trying to unify GR with quantum effects ..for instance am stating that divergences at zero point in GR that appear to be non renormalizable are just normal issues of quantum fluctuations that are as a result of rich world inside a graviton....if we equate all this to information, graviton represent the threshold at which this information(hidden curvature) inside the graviton leaks to the outside curvature that becomes detectable by GR i.e a point in the manifold.

I’m afraid this doesn’t make any sense.

17 hours ago, Markus Hanke said:

the Lie bracket of two vector fields measures the extent to which differential operators associated with these fields fail to commute; specifically in this case, it’s related to the commutator of covariant derivatives, which on a Riemann manifold is not zero, unless the manifold is perfectly flat.

Am suggesting a situation where when parallel transporting a vector in this case on a parallelogram loop lie bracket becomes zero meaning the manifold is perfectly flat...yet it's due to uniformity of information between the end points of the two vectors of the parallelogram because the rate at which we are obtaining this information is limited to the speed of light...am trying to look for a better way to explain this..

4 hours ago, MJ kihara said:

Am suggesting a situation where when parallel transporting a vector in this case on a parallelogram loop lie bracket becomes zero meaning the manifold is perfectly flat...

Ok, so you’re working on a flat manifold.

But what do you mean by “loop Lie bracket”? What are you taking a Lie bracket of, exactly?

4 hours ago, MJ kihara said:

yet it's due to uniformity of information between the end points of the two vectors of the parallelogram because the rate at which we are obtaining this information is limited to the speed of light...am trying to look for a better way to explain this..

I don’t know what “uniformity of information” means.

If the manifold is flat, as you stated above, then the vectors will coincide after parallel transport; but of course that means you don’t have any gravity in this situation.

9 hours ago, Markus Hanke said:

Ok, so you’re working on a flat manifold.

Yes,a manifold perceived to be flat.

10 hours ago, Markus Hanke said:

But what do you mean by “loop Lie bracket”? What are you taking a Lie bracket of, exactly?

In my assumptions am using a parallelogram loop.

The lie bracket am talking about,simply defined as per Wikipedia... the Lie bracket of vector fields, also known as the Jacobi–Lie bracket or the commutator of vector fields, is an operator that assigns to any two vector fields X and Y on a smooth manifold M third vector field denoted [X,Y].

The gap between the two basis vector of parallelogram to make a complete loop for a smooth movement of parallel transported vector on a flat manifold it's taken to be zero as per Riemann tensor.

11 hours ago, Markus Hanke said:

I don’t know what “uniformity of information” means.

What I mean is the smooth movement without detecting a gap [X,Y]=0...the 'smoothness' is the one am referring to as uniformity of information,that is,information of X axis side of parallelogram to Y axis side of the parallelogram...assuming the vector being parallel transported is aware, it could not know it has been transported from one basis vector(sides of the parallelogram) to another, the information its gaining is uniform.

11 hours ago, Markus Hanke said:

If the manifold is flat, as you stated above, then the vectors will coincide after parallel transport; but of course that means you don’t have any gravity in this situation.

If it's a point in the manifold Christoffel symbols reduces to zero and the metric tensor becomes the identity matrix if am not wrong.hope the idea am trying to present make sense.....this point is part of the curvature of the manifold....to visualize...if the Riemann tensor is able to detect just the point of this manifold without gaining further information i.e it's just sliding on the tangent plane of the manifold....to it,the manifold will just be flat while there is the 'hidden curvature' that is the rest of the manifold.

This point that am explaining about,is the one am talking about being a point between the the basis vector of the parallelogram when it closes..if I were to use a drawing to illustrate this I would draw a parallelogram and at the further end where basis vector close,at that point,draw a circle below it.

13 hours ago, MJ kihara said:

The gap between the two basis vector of parallelogram to make a complete loop for a smooth movement of parallel transported vector on a flat manifold it's taken to be zero as per Riemann tensor.

Yes, so long as the manifold is flat.

13 hours ago, MJ kihara said:

If it's a point in the manifold Christoffel symbols reduces to zero and the metric tensor becomes the identity matrix if am not wrong.

No they don’t, unless the manifold is flat. All tensors are purely local objects, and the metric tensor in particular allows you to define the notion of inner product at a point. However, the numerical values of the components of the tensor depend on where they are evaluated (thus the Christoffel symbols don’t automatically vanish), unless the manifold is flat. And then the metric tensor becomes the Minkowski metric, since we are working in GR.

13 hours ago, MJ kihara said:

if the Riemann tensor is able to detect just the point of this manifold

The Riemann tensor is defined at a point, like all tensors, but it is made up of derivatives of the metric, which don’t vanish on a curved manifold.

All these tensors - the metric, Riemann etc - are in fact tensor fields, ie they are not isolated objects, but defined at all points of the manifold. So there’s no such thing as “hidden curvature” - a manifold is either curved, so that Riemann doesn’t vanish, or it isn’t, in which case Riemann is zero. There are no other options here.

Note that all of this is very precisely defined using calculus on manifolds, and well understood.

10 hours ago, Markus Hanke said:

No they don’t, unless the manifold is flat. All tensors are purely local objects, and the metric tensor in particular allows you to define the notion of inner product at a point.

Am not disputing this fact..using Riemann normal coordinate at point p the metric tensor becomes the identity matrix and Christoffel symbols equals zero...that is the point am referring to... isn't used to prove second Bianchi identity?

10 hours ago, Markus Hanke said:

All these tensors - the metric, Riemann etc - are in fact tensor fields, ie they are not isolated objects, but defined at all points of the manifold. So there’s no such thing as “hidden curvature” - a manifold is either curved, so that Riemann doesn’t vanish, or it isn’t, in which case Riemann is zero. There are no other options here.

Note that all of this is very precisely defined using calculus on manifolds, and well understood.

Am not disputing the established math am just trying to use it to achieve what am thinking.

If you evaluate a straight line on top of a circle ...you won't detect the curvature of the circle below the line using Riemann tensor,as you have said,it reduces to Mankowski metric unless you are aware there is a circle...e.g you put an ant and restrict it to move forward either way on the line it won't realize there is a circle below it,the circle is the example of a hidden curvature..where the circle meets the line becomes just a point in the line..hope am not wrong with that reasoning.

52 minutes ago, MJ kihara said:

Am not disputing this fact..using Riemann normal coordinate at point p the metric tensor becomes the identity matrix and Christoffel symbols equals zero...that is the point am referring to... isn't used to prove second Bianchi identity?

Normal coordinates can be used to prove Riemann tensor identities such as the second Bianchi identity, but they need not be used to prove such identities. I prefer not to use normal coordinates to prove such identities, but to prove them directly.

1 hour ago, MJ kihara said:

If you evaluate a straight line on top of a circle ...you won't detect the curvature of the circle below the line using Riemann tensor,as you have said,it reduces to Mankowski metric unless you are aware there is a circle...e.g you put an ant and restrict it to move forward either way on the line it won't realize there is a circle below it,the circle is the example of a hidden curvature..where the circle meets the line becomes just a point in the line..hope am not wrong with that reasoning.

Even though the Christoffel symbols are zero in normal coordinates, the partial derivatives of the Christoffel symbols are generally not zero and the Riemann tensor is generally not zero. The Christoffel symbols are not a tensor whereas the Riemann tensor is a tensor. Thus, whereas the Christoffel symbols can be coordinate-transformed to zero at a point, the Riemann tensor in general can't be coordinate-transformed to zero at the point.

The Riemann tensor may be defined in terms of parallel transport of vectors around infinitesimal parallelograms. Although the resulting change in the vector is itself infinitesimal, the Riemann tensor that is obtained as a result is not infinitesimal.

Edited May 15May 15 by KJW

4 hours ago, KJW said:

Normal coordinates can be used to prove Riemann tensor identities such as the second Bianchi identity, but they need not be used to prove such identities. I prefer not to use normal coordinates to prove such identities, but to prove them directly.

All the same they simplify the math and leads to the correct answer.

4 hours ago, KJW said:

Even though the Christoffel symbols are zero in normal coordinates, the partial derivatives of the Christoffel symbols are generally not zero and the Riemann tensor is generally not zero.

Derivative of Christoffel symbols in one direction reduces to zero...when then done in another extra direction it doesn't vanish.....this appears to indicate that when the degree of freedom increases it become more sensitive in extracting information from the manifold, I'm basing this argument on proving second Bianchi identity at a point.

5 hours ago, KJW said:

Thus, whereas the Christoffel symbols can be coordinate-transformed to zero at a point, the Riemann tensor in general can't be coordinate-transformed to zero at the point.

By introducing some constraints, and cycling the lower indices adding up the available degrees of freedom Riemann tensor in a given direction reduces to zero as indicated in the second Bianchi identity.

5 hours ago, KJW said:

The Riemann tensor may be defined in terms of parallel transport of vectors around infinitesimal parallelograms. Although the resulting change in the vector is itself infinitesimal, the Riemann tensor that is obtained as a result is not infinitesimal.

What is the importance of lie bracket in all of this,why bother adding it up in Riemann tensor formula?

13 hours ago, MJ kihara said:

e.g you put an ant and restrict it to move forward either way on the line it won't realize there is a circle below it

Sure it will, because eventually the ant will end up again at the same point where it started, and thus realise that it wasn’t on a flat surface. While its trajectory was everywhere locally flat, the global geodesic it followed was a circle.

13 hours ago, MJ kihara said:

isn't used to prove second Bianchi identity?

Sure you can use these to prove the Bianchi identity, but you don’t have to. As KJW has pointed out already, you must remember that the Christoffel symbols aren’t tensors, so you can always make coordinate choices such that they locally vanish. However, Riemann is a tensor, so an a curved surface it can never be made to vanish.

6 hours ago, MJ kihara said:

What is the importance of lie bracket in all of this,why bother adding it up in Riemann tensor formula?

It measures the commutativity of the covariant derivative. On curved surfaces, covariant derivatives do not commute, and the extent to which they fail to do so is measured by the Riemann tensor.

10 hours ago, Markus Hanke said:

Sure it will, because eventually the ant will end up again at the same point where it started, and thus realise that it wasn’t on a flat surface. While its trajectory was everywhere locally flat, the global geodesic it followed was a circle.

It's seems we are getting each other wrong...to be clear make a very(infinitely )thin wire into a circle,then another one flat(straight) which extend, let's say infinitely...put the straight(flat) wire on top of the circle then put an ant on it then restrict it to move on forward or backwards...how will it detect the circle below?

It's moving on the straight wire.

12 hours ago, MJ kihara said:

It's seems we are getting each other wrong...to be clear make a very(infinitely )thin wire into a circle,then another one flat(straight) which extend, let's say infinitely...put the straight(flat) wire on top of the circle then put an ant on it then restrict it to move on forward or backwards...how will it detect the circle below?

It's moving on the straight wire.

I don’t know what this is supposed to show, to be honest. One wire is infinite, the other one is not (it returns to the same starting point). In both cases they look locally flat. But globally these situations differ.

In GR, any small enough patch of spacetime can locally be considered Minkowski; but this doesn’t at all imply that the entire spacetime is globally without curvature, or that curvature is somehow “hidden”, it just means that curvature effects are small enough to become negligible, but they are still there. For example, longitudinal lines on Earth still converge at the poles, even if locally on a small enough scale they appear to be parallel.

This is where the connection (and thus the Riemann tensor) come in, because it tells is how these local patches are globally related to one another.

On 5/16/2025 at 8:36 AM, MJ kihara said:

On 5/16/2025 at 3:12 AM, KJW said:

Normal coordinates can be used to prove Riemann tensor identities such as the second Bianchi identity, but they need not be used to prove such identities. I prefer not to use normal coordinates to prove such identities, but to prove them directly.

All the same they simplify the math and leads to the correct answer.

That depends on the connection. For a Riemannian connection, normal coordinates exist which simplify the proofs of identities. But for a general connection, for example if the torsion tensor is non-zero, then normal coordinates do not exist in general. Also, if one is using normal coordinates to simplify the proofs of Riemann tensor identities, then one also needs to include the proof of the existence of the normal coordinates themselves. Thus, in the end it may actually be simpler to prove the Riemann tensor identities directly without invoking normal coordinates. And while one is at it, one may as well prove the Riemann tensor identities for the general connection.