Tensors and General Covariance

[Alan Grayson]

In some treatments of tensors, they're described as linear maps. So, in GR, if we have a linear map described as a 4x4 matrix of real numbers, which operates on a 4-vector described as a column matrix with entries (ct, x, y, z), which transforms to another 4-vector, what must be added in this description to claim that the linear transformation satisfies the definition of a tensor? TY, AG

[Alan Grayson]

On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote:

In some treatments of tensors, they're described as linear maps. So, in GR, if we have a linear map described as a 4x4 matrix of real numbers, which operates on a 4-vector described as a column matrix with entries (ct, x, y, z), which transforms to another 4-vector, what must be added in this description to claim that the linear transformation satisfies the definition of a tensor? TY, AG

Let's call the linear transformation T, then the answer to my question might be that T is a tensor iff it has a continuous inverse.  I'm not sure if this is correct, but I seem to recall this claim in a video about tensors I viewed in another life. But even if it's true, it seems to conflict with the claim that an ordinary vector in Euclidean space is a tensor because it's invariant under linear (?) transformations. In this formulation, it is the argument of T, which we can call V, which is invariant, not the map T. I'd appreciate it if someone here could clarity my confusion. TY, AG

[Brent Meeker]

A tensor is a geometric object (possibly in an abstract space).  It transforms covariantly; which means that changes in coordinates (even non-linear changes in coordinates) leave it the same.  A physical vector, like velocity, is a one-dimensional tensor.  The same vector, or tensor, has different representations in different coordinate systems.  Here's an excerpt from my general relativity lectures.

Vectors are used to represent things like motion, force, flow. You think of them as little arrows that show the direction of the motion, force or flow and the length of the arrow tells you the magnitude ot the motion, force or flow. So vectors have terms, one for each dimension of the space they are in. In the plane, which is two dimensional, vectors have two components. But the values of the components change depending on which coordinate system you choose to represent them. The coordinates are something we impose to facilitate our calculation. But the vector or tensor is a THING that is independent of the coordinates we use to describe them. Just as I could give directions from this building to the upper parking lot by saying it's 250ft that way or I could say it's 200ft north and 150ft west. The distance and direction would be the same, only the description is different.

Tensors are just a another step up from vectors. They describe how vectors are changing. Here's a good example of a vector field. It's called a field because there's a vector at each point. It shows the flow of water out of Monterey bay at a particular moment. 

For any particular flow field we can ask how do the vectors around a particular bit of water change as that bit of water is carried along by the flow. The answer is a tensor. Just as a vector is the abstraction of a little arrow, the tensor can be thought of an as abstraction of a little circle. And then the question becomes how does this circle get distorted as the water flows.

If the water is moving uniformly then the circle doesn't get distorted.  This is like the no-curvature tensor.  But the flow in Monterey Bay is not uniform, so the circle/tensor gets distorted.  

For flow like this in two dimensions the tensor is just an ellipse.  It has two directions corresponding to the axes of the ellipse and it has a size corresponding to the strength of the flow.  So it only takes three number to describe it.  The Txy component is the same as the Tyx component.   But this is a tensor FIELD. So there's a different tensor at each point.  It takes 3 numbers at each point.

If we choose some particular coordinate system then the components have interpretations like, “How much does the x flow speed change as you move in the y direction.”

[Russell Standish]

It's got nothing to do with being invertible (which is the conjunction
of being 1:1 and onto).

Rather a tensor is a multilinear map - is a map with multiple
arguments, and linear in each. Obviously a standard linear map R^n -> R^n is a
rank 2 tensor. We recognise them generally as matrices. Vectors
correspond to linear maps by means of transposing them and forming the
inner product, ie a linear map from R^n->R, and are rank 1 tensors as a result.



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--

----------------------------------------------------------------------------
Dr Russell Standish Phone 0425 253119 (mobile)
Principal, High Performance Coders hpc...@hpcoders.com.au
http://www.hpcoders.com.au
----------------------------------------------------------------------------

[Alan Grayson]

On Sunday, November 9, 2025 at 2:07:30 PM UTC-7 Brent Meeker wrote:

On 11/9/2025 1:11 AM, Alan Grayson wrote:

On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote:

In some treatments of tensors, they're described as linear maps. So, in GR, if we have a linear map described as a 4x4 matrix of real numbers, which operates on a 4-vector described as a column matrix with entries (ct, x, y, z), which transforms to another 4-vector, what must be added in this description to claim that the linear transformation satisfies the definition of a tensor? TY, AG

Let's call the linear transformation T, then the answer to my question might be that T is a tensor iff it has a continuous inverse.  I'm not sure if this is correct, but I seem to recall this claim in a video about tensors I viewed in another life. But even if it's true, it seems to conflict with the claim that an ordinary vector in Euclidean space is a tensor because it's invariant under linear (?) transformations. In this formulation, it is the argument of T, which we can call V, which is invariant, not the map T. I'd appreciate it if someone here could clarity my confusion. TY, AG

A tensor is a geometric object (possibly in an abstract space).  It transforms covariantly; which means that changes in coordinates (even non-linear changes in coordinates) leave it the same. 

Round and round we go, but what a tensor is remains elusive! Please define the property that allows it to transform covariantly. Is it a map represented by a matix, and if so, what property must it have that allows it to transform covariantly? AG

[Alan Grayson]

On Sunday, November 9, 2025 at 3:06:45 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 01:11:47AM -0800, Alan Grayson wrote:
>
>
> On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote:
>
> In some treatments of tensors, they're described as linear maps. So, in GR,
> if we have a linear map described as a 4x4 matrix of real numbers, which
> operates on a 4-vector described as a column matrix with entries (ct, x, y,
> z), which transforms to another 4-vector, what must be added in this
> description to claim that the linear transformation satisfies the
> definition of a tensor? TY, AG
>
>
> Let's call the linear transformation T, then the answer to my question might be
> that T is a tensor iff it has a continuous inverse.  I'm not sure if this is
> correct, but I seem to recall this claim in a video about tensors I viewed in
> another life. But even if it's true, it seems to conflict with the claim that
> an ordinary vector in Euclidean space is a tensor because it's invariant under
> linear (?) transformations. In this formulation, it is the argument of T, which
> we can call V, which is invariant, not the map T. I'd appreciate it if someone
> here could clarity my confusion. TY, AG

It's got nothing to do with being invertible (which is the conjunction
of being 1:1 and onto).

Rather a tensor is a multilinear map -

If it's a map, how can an ordinary vector in Euclidean space be a tensor?

Such vectors are NOT maps! See my problem? AG

 

is a map with multiple
arguments, and linear in each. Obviously a standard linear map R^n -> R^n is a
rank 2 tensor. We recognise them generally as matrices. Vectors
correspond to linear maps by means of transposing them and forming the
inner product, ie a linear map from R^n->R, and are rank 1 tensors as a result.



>

[Russell Standish]

On Sun, Nov 09, 2025 at 03:56:16PM -0800, Alan Grayson wrote:
>
> If it's a map, how can an ordinary vector in Euclidean space be a tensor?
> Such vectors are NOT maps! See my problem? AG

I did explain that in my post if you read it. In an inner product
space, every vector is isomorphic to a linear map from the space to
its field. Eg R^n->R in the case of the space R^n. That linear map is
the rank 1 tensor. In mathematics, something walks and quacks like a
duck is a duck.

Even the inner product operation is an example of a bilinear map,
hence a rank 2 tensor. In Minkowski spacetime, the inner product is
known as the Levi-Civita tensor.


--

----------------------------------------------------------------------------

[Alan Grayson]

On Sunday, November 9, 2025 at 5:12:54 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 03:56:16PM -0800, Alan Grayson wrote:
>
> If it's a map, how can an ordinary vector in Euclidean space be a tensor?
> Such vectors are NOT maps! See my problem? AG


I did explain that in my post if you read it. In an inner product
space, every vector is isomorphic to a linear map from the space to
its field. Eg R^n->R in the case of the space R^n. That linear map is
the rank 1 tensor. In mathematics, something walks and quacks like a
duck is a duck.

Even the inner product operation is an example of a bilinear map,
hence a rank 2 tensor. In Minkowski spacetime, the inner product is
known as the Levi-Civita tensor.

So a tensor is nothing more than a multi linear map to the reals? But if

we represent a tensor by a matrix, will it be automatically invariant 

under coordinate transformations? Do we need an inner product space

to define a tensor? TY, AG 

[Alan Grayson]

On Sunday, November 9, 2025 at 6:16:15 PM UTC-7 Alan Grayson wrote:

On Sunday, November 9, 2025 at 5:12:54 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 03:56:16PM -0800, Alan Grayson wrote:
>
> If it's a map, how can an ordinary vector in Euclidean space be a tensor?
> Such vectors are NOT maps! See my problem? AG


I did explain that in my post if you read it. In an inner product
space, every vector is isomorphic to a linear map from the space to
its field. Eg R^n->R in the case of the space R^n. That linear map is
the rank 1 tensor. In mathematics, something walks and quacks like a
duck is a duck.

Even the inner product operation is an example of a bilinear map,
hence a rank 2 tensor. In Minkowski spacetime, the inner product is
known as the Levi-Civita tensor.

So a tensor is nothing more than a multi linear map to the reals? But if

we represent a tensor by a matrix, will it be automatically invariant 

under coordinate transformations? Do we need an inner product space

to define a tensor? TY, AG 

If the tensor, represented by a matrix, is "unchanged" under a coordinate

transformation, does this mean its determinant is unchanged? AG 

[Brent Meeker]

On 11/9/2025 3:49 PM, Alan Grayson wrote:

On Sunday, November 9, 2025 at 2:07:30 PM UTC-7 Brent Meeker wrote:

On 11/9/2025 1:11 AM, Alan Grayson wrote:

On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote:

In some treatments of tensors, they're described as linear maps. So, in GR, if we have a linear map described as a 4x4 matrix of real numbers, which operates on a 4-vector described as a column matrix with entries (ct, x, y, z), which transforms to another 4-vector, what must be added in this description to claim that the linear transformation satisfies the definition of a tensor? TY, AG

Let's call the linear transformation T, then the answer to my question might be that T is a tensor iff it has a continuous inverse.  I'm not sure if this is correct, but I seem to recall this claim in a video about tensors I viewed in another life. But even if it's true, it seems to conflict with the claim that an ordinary vector in Euclidean space is a tensor because it's invariant under linear (?) transformations. In this formulation, it is the argument of T, which we can call V, which is invariant, not the map T. I'd appreciate it if someone here could clarity my confusion. TY, AG

A tensor is a geometric object (possibly in an abstract space).  It transforms covariantly; which means that changes in coordinates (even non-linear changes in coordinates) leave it the same. 

Round and round we go, but what a tensor is remains elusive! Please define the property that allows it to transform covariantly. Is it a map represented by a matix, and if so, what property must it have that allows it to transform covariantly? AG

So you just read far enough to think of a question. 

[Alan Grayson]

BS! It's universally claimed that tensors remain unchanged under coordinate transformations, but rarely, if ever, is the property tensors must have to have this result. Your plot is no different. Does its matrix representation remain unchanged? Does this MEAN its determinant is unchanged? If not, please specify the property and test for this invariant. AG 

[Brent Meeker]

On 11/9/2025 5:24 PM, Alan Grayson wrote:

On Sunday, November 9, 2025 at 6:16:15 PM UTC-7 Alan Grayson wrote:

On Sunday, November 9, 2025 at 5:12:54 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 03:56:16PM -0800, Alan Grayson wrote:
>
> If it's a map, how can an ordinary vector in Euclidean space be a tensor?
> Such vectors are NOT maps! See my problem? AG


I did explain that in my post if you read it. In an inner product
space, every vector is isomorphic to a linear map from the space to
its field. Eg R^n->R in the case of the space R^n. That linear map is
the rank 1 tensor. In mathematics, something walks and quacks like a
duck is a duck.

Even the inner product operation is an example of a bilinear map,
hence a rank 2 tensor. In Minkowski spacetime, the inner product is
known as the Levi-Civita tensor.

So a tensor is nothing more than a multi linear map to the reals? But if

we represent a tensor by a matrix, will it be automatically invariant 

under coordinate transformations? Do we need an inner product space

to define a tensor? TY, AG 

If the tensor, represented by a matrix, is "unchanged" under a coordinate

transformation, does this mean its determinant is unchanged? AG 

No, in general it transforms like a density.  So it's only unchanged if the determinant of the transformation matrix is 1 

Brent

--

----------------------------------------------------------------------------
Dr Russell Standish Phone 0425 253119 (mobile)
Principal, High Performance Coders hpc...@hpcoders.com.au
http://www.hpcoders.com.au
----------------------------------------------------------------------------

--

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[Alan Grayson]

On Sunday, November 9, 2025 at 8:01:50 PM UTC-7 Brent Meeker wrote:

On 11/9/2025 5:24 PM, Alan Grayson wrote:

On Sunday, November 9, 2025 at 6:16:15 PM UTC-7 Alan Grayson wrote:

On Sunday, November 9, 2025 at 5:12:54 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 03:56:16PM -0800, Alan Grayson wrote:
>
> If it's a map, how can an ordinary vector in Euclidean space be a tensor?
> Such vectors are NOT maps! See my problem? AG


I did explain that in my post if you read it. In an inner product
space, every vector is isomorphic to a linear map from the space to
its field. Eg R^n->R in the case of the space R^n. That linear map is
the rank 1 tensor. In mathematics, something walks and quacks like a
duck is a duck.

Even the inner product operation is an example of a bilinear map,
hence a rank 2 tensor. In Minkowski spacetime, the inner product is
known as the Levi-Civita tensor.

So a tensor is nothing more than a multi linear map to the reals? But if

we represent a tensor by a matrix, will it be automatically invariant 

under coordinate transformations? Do we need an inner product space

to define a tensor? TY, AG 

If the tensor, represented by a matrix, is "unchanged" under a coordinate

transformation, does this mean its determinant is unchanged? AG 

No, in general it transforms like a density.  So it's only unchanged if the determinant of the transformation matrix is 1 

Brent

Someday I might find a teacher who can really define tensors, but that day has yet to arrive. Standish seems to come close, but does every linear multivariate function define a tensor? I'm waiting to see his reply. AG

[Russell Standish]

On Sun, Nov 09, 2025 at 09:55:15PM -0800, Alan Grayson wrote:
>
>
> Someday I might find a teacher who can really define tensors, but that day has
> yet to arrive. Standish seems to come close, but does every linear multivariate
> function define a tensor? I'm waiting to see his reply. AG

Well I did say multilinear function, but the answer is yes, every
multilinear function on a vector space is a tensor, and vice-versa.

I did write an 8 page article appearing in our student rag "The
Occasional Quark" when I was a physics student, which was my attempt
at explaining General Relativity when I was disgusted by the hash job
done by our professor. I haven't really thought about it much since
that time, though. I can also recommend the heavy tome by Misner,
Thorne and Wheeler.

I could scan the article and post it to this list, but not today - I
have a few other things on my plate before finishing up.

[Alan Grayson]

On Sunday, November 9, 2025 at 11:16:05 PM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 09:55:15PM -0800, Alan Grayson wrote:
>
>
> Someday I might find a teacher who can really define tensors, but that day has
> yet to arrive. Standish seems to come close, but does every linear multivariate
> function define a tensor? I'm waiting to see his reply. AG

Well I did say multilinear function, but the answer is yes, every
multilinear function on a vector space is a tensor, and vice-versa.

How does one prove that every multilinear function on a vector space is invariant under a 

change in coordinates? What exactly happens to its matrix representation? And Yes, please

post that short clarification defining tensors when you have time. AG

[Russell Standish]

On Sun, Nov 09, 2025 at 10:34:38PM -0800, Alan Grayson wrote:
>
>
> On Sunday, November 9, 2025 at 11:16:05 PM UTC-7 Russell Standish wrote:
>
> On Sun, Nov 09, 2025 at 09:55:15PM -0800, Alan Grayson wrote:
> >
> >
> > Someday I might find a teacher who can really define tensors, but that
> day has
> > yet to arrive. Standish seems to come close, but does every linear
> multivariate
> > function define a tensor? I'm waiting to see his reply. AG
>
> Well I did say multilinear function, but the answer is yes, every
> multilinear function on a vector space is a tensor, and vice-versa.
>
>
> How does one prove that every multilinear function on a vector space is
> invariant under a 
> change in coordinates? What exactly happens to its matrix representation? And

A vector is a geometric quantity having direction and length. As such
it is independent of any coordinate system that might be applied to
the space (although the list of numbers representing the components of
the vector in a given coordinate system must vary covariantly with the
coordinate system varying). A function operating on vectors, and
returning vectors or scalars must therfore also be independent of the
coordinate system.

> Yes, please
> post that short clarification defining tensors when you have time. AG


>
>
> I did write an 8 page article appearing in our student rag "The
> Occasional Quark" when I was a physics student, which was my attempt
> at explaining General Relativity when I was disgusted by the hash job
> done by our professor. I haven't really thought about it much since
> that time, though. I can also recommend the heavy tome by Misner,
> Thorne and Wheeler.
>
> I could scan the article and post it to this list, but not today - I
> have a few other things on my plate before finishing up.
>
>
>
> --
>
> ----------------------------------------------------------------------------
>
> Dr Russell Standish Phone 0425 253119 (mobile)
> Principal, High Performance Coders hpc...@hpcoders.com.au
> http://www.hpcoders.com.au
> ----------------------------------------------------------------------------
>
>

> --
> You received this message because you are subscribed to the Google Groups
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> To unsubscribe from this group and stop receiving emails from it, send an email
> to everything-li...@googlegroups.com.
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[Alan Grayson]

On Sunday, November 9, 2025 at 2:07:30 PM UTC-7 Brent Meeker wrote:

On 11/9/2025 1:11 AM, Alan Grayson wrote:

On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote:

In some treatments of tensors, they're described as linear maps. So, in GR, if we have a linear map described as a 4x4 matrix of real numbers, which operates on a 4-vector described as a column matrix with entries (ct, x, y, z), which transforms to another 4-vector, what must be added in this description to claim that the linear transformation satisfies the definition of a tensor? TY, AG

Let's call the linear transformation T, then the answer to my question might be that T is a tensor iff it has a continuous inverse.  I'm not sure if this is correct, but I seem to recall this claim in a video about tensors I viewed in another life. But even if it's true, it seems to conflict with the claim that an ordinary vector in Euclidean space is a tensor because it's invariant under linear (?) transformations. In this formulation, it is the argument of T, which we can call V, which is invariant, not the map T. I'd appreciate it if someone here could clarity my confusion. TY, AG

A tensor is a geometric object (possibly in an abstract space). 

It's an ALGEBRAIC object, NOT a geometric object, defined on a vector space, as a multilinear function which maps to a real number. The inner product function on an inner product space is a simple example of a tensor. To prove it's invariant under a change in coordinates, one must take two arbitrary vectors in the vector space under consideration, perform a change of coordinates, and prove the mapping of the function remains unchanged. When I figure out how that's done, I'll post it. AG

[Russell Standish]

On Mon, Nov 10, 2025 at 12:19:49AM -0800, Alan Grayson wrote:
>
> It's an ALGEBRAIC object, NOT a geometric object, defined on a vector space, as
> a multilinear function which maps to a real number.

Calling it an _algebraic_ object is exactly what obfuscates what
tensors are all about. Tensors are not a collection of numbers, just
as vectors are not collections of numbers. Vectors are geometric
objects, as are tensors.

[Alan Grayson]

On Monday, November 10, 2025 at 1:15:42 AM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 10:34:38PM -0800, Alan Grayson wrote:
>
>
> On Sunday, November 9, 2025 at 11:16:05 PM UTC-7 Russell Standish wrote:
>
> On Sun, Nov 09, 2025 at 09:55:15PM -0800, Alan Grayson wrote:
> >
> >
> > Someday I might find a teacher who can really define tensors, but that
> day has
> > yet to arrive. Standish seems to come close, but does every linear
> multivariate
> > function define a tensor? I'm waiting to see his reply. AG
>
> Well I did say multilinear function, but the answer is yes, every
> multilinear function on a vector space is a tensor, and vice-versa.
>
>
> How does one prove that every multilinear function on a vector space is
> invariant under a 
> change in coordinates? What exactly happens to its matrix representation? And

A vector is a geometric quantity having direction and length. As such
it is independent of any coordinate system that might be applied to
the space (although the list of numbers representing the components of
the vector in a given coordinate system must vary covariantly with the
coordinate system varying). A function operating on vectors, and
returning vectors or scalars must therfore also be independent of the
coordinate system.

This is OK, because a tensor is defined for vectors which are invariant 

under changes in coordinates. AG

[Alan Grayson]

On Monday, November 10, 2025 at 2:05:39 AM UTC-7 Russell Standish wrote:

On Mon, Nov 10, 2025 at 12:19:49AM -0800, Alan Grayson wrote:
>
> It's an ALGEBRAIC object, NOT a geometric object, defined on a vector space, as
> a multilinear function which maps to a real number.

Calling it an _algebraic_ object is exactly what obfuscates what
tensors are all about. Tensors are not a collection of numbers, just
as vectors are not collections of numbers. Vectors are geometric
objects, as are tensors.

I disagree. Vectors are geometric objects, but tensors are algebraic in

that they're defined as functions. AG 

[Alan Grayson]

On Sunday, November 9, 2025 at 2:07:30 PM UTC-7 Brent Meeker wrote:

On 11/9/2025 1:11 AM, Alan Grayson wrote:

On Saturday, November 8, 2025 at 6:25:17 AM UTC-7 Alan Grayson wrote:

In some treatments of tensors, they're described as linear maps. So, in GR, if we have a linear map described as a 4x4 matrix of real numbers, which operates on a 4-vector described as a column matrix with entries (ct, x, y, z), which transforms to another 4-vector, what must be added in this description to claim that the linear transformation satisfies the definition of a tensor? TY, AG

Let's call the linear transformation T, then the answer to my question might be that T is a tensor iff it has a continuous inverse.  I'm not sure if this is correct, but I seem to recall this claim in a video about tensors I viewed in another life. But even if it's true, it seems to conflict with the claim that an ordinary vector in Euclidean space is a tensor because it's invariant under linear (?) transformations. In this formulation, it is the argument of T, which we can call V, which is invariant, not the map T. I'd appreciate it if someone here could clarity my confusion. TY, AG

A tensor is a geometric object (possibly in an abstract space).  It transforms covariantly; which means that changes in coordinates (even non-linear changes in coordinates) leave it the same.  A physical vector, like velocity, is a one-dimensional tensor.  The same vector, or tensor, has different representations in different coordinate systems.  Here's an excerpt from my general relativity lectures.

Vectors are used to represent things like motion, force, flow. You think of them as little arrows that show the direction of the motion, force or flow and the length of the arrow tells you the magnitude ot the motion, force or flow. So vectors have terms, one for each dimension of the space they are in. In the plane, which is two dimensional, vectors have two components. But the values of the components change depending on which coordinate system you choose to represent them. The coordinates are something we impose to facilitate our calculation. But the vector or tensor is a THING that is independent of the coordinates we use to describe them. Just as I could give directions from this building to the upper parking lot by saying it's 250ft that way or I could say it's 200ft north and 150ft west. The distance and direction would be the same, only the description is different.

Tensors are just a another step up from vectors. They describe how vectors are changing. Here's a good example of a vector field. It's called a field because there's a vector at each point. It shows the flow of water out of Monterey bay at a particular moment. 

For any particular flow field we can ask how do the vectors around a particular bit of water change as that bit of water is carried along by the flow. The answer is a tensor. Just as a vector is the abstraction of a little arrow, the tensor can be thought of an as abstraction of a little circle. And then the question becomes how does this circle get distorted as the water flows.

Since a tensor maps to real numbers, the tensor field consists of real numbers, not as you have it, as vectors. AG 

[Alan Grayson]

On Monday, November 10, 2025 at 2:05:39 AM UTC-7 Russell Standish wrote:

On Mon, Nov 10, 2025 at 12:19:49AM -0800, Alan Grayson wrote:
>
> It's an ALGEBRAIC object, NOT a geometric object, defined on a vector space, as
> a multilinear function which maps to a real number.

Calling it an _algebraic_ object is exactly what obfuscates what
tensors are all about. Tensors are not a collection of numbers, just
as vectors are not collections of numbers. Vectors are geometric
objects, as are tensors.

A tensor field is, in fact, a collection of real numbers at different 

positions in the field. AG 

[Brent Meeker]

Why don't you read a book?  Russell gave you a definition.  I gave you an example.  Nobody wants to write a lot of math text online.

Brent

[Alan Grayson]

Why don't you read my comments before replying? I accept Russell's definiiton. Moreover, a tensor can be an inner product, and maps to a real numbers, so a tensor field is not like many arrows but real numbers. So your example is misleading. Like most teachers of tensors, you are averse to giving a precise definition, which Russell did. AG  

[Alan Grayson]

You write "Nobody wants to write a lot of math text online." Do you know that the proof that tensors are invariant to changes in coordinate systems does NOT involve writing any mathematics? It follows directly from the definition of tensors. AG

[Alan Grayson]

On Monday, November 10, 2025 at 1:15:42 AM UTC-7 Russell Standish wrote:

On Sun, Nov 09, 2025 at 10:34:38PM -0800, Alan Grayson wrote:
>
>
> On Sunday, November 9, 2025 at 11:16:05 PM UTC-7 Russell Standish wrote:
>
> On Sun, Nov 09, 2025 at 09:55:15PM -0800, Alan Grayson wrote:
> >
> >
> > Someday I might find a teacher who can really define tensors, but that
> day has
> > yet to arrive. Standish seems to come close, but does every linear
> multivariate
> > function define a tensor? I'm waiting to see his reply. AG
>
> Well I did say multilinear function, but the answer is yes, every
> multilinear function on a vector space is a tensor, and vice-versa.
>
>
> How does one prove that every multilinear function on a vector space is
> invariant under a 
> change in coordinates? What exactly happens to its matrix representation? And

A vector is a geometric quantity having direction and length. As such
it is independent of any coordinate system that might be applied to
the space (although the list of numbers representing the components of
the vector in a given coordinate system must vary covariantly with the
coordinate system varying). A function operating on vectors, and
returning vectors or scalars must therfore also be independent of the
coordinate system.

I think you deserve a special "THANK YOU!" Now, after many years, I finally

understand why tensors are invariant under changes in coordinate systems.

The problem has always been the reluctance or inability for most physicists

to explicity define the MATHEMATICAL definition of a tensor. They usually omit this

necessary definition by just asserting they're mathematical entities which

are invariant under changes in coordinate systems. The explanation is totally

simple; namely, tensors are defined on vectors in vector spaces, and these

vectors are invariant under changes in coordinate systems, so the function

defining tensors must likewise have this property since these vectors are

the domains on which these functions operate. NOT COMPLICATED! I think

you nailed it because your core orientation is more in mathematics than physics. 

Finally, please post your summary of GR when you have time. I am anxious to

read it. AG

> I did write an 8 page article appearing in our student rag "The
> Occasional Quark" when I was a physics student, which was my attempt
> at explaining General Relativity when I was disgusted by the hash job
> done by our professor. I haven't really thought about it much since
> that time, though. I can also recommend the heavy tome by Misner,
> Thorne and Wheeler.
>
> I could scan the article and post it to this list, but not today - I
> have a few other things on my plate before finishing up.
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