Covariant and Contravariant: What Are the Differences in Differential Geometry?

[StatusX]

I'm starting to learn differential geometetry on my own, but I'm having a little trouble figuring out the difference between covariant and contravariant vector fields. It seems that contravariant fields are just the normal vector fields they introduced in multivariable calculus, but if so, I can't figure out what covariant fields are.

 

[dextercioby]

I'm starting to learn differential geometetry on my own, but I'm having a little trouble figuring out the difference between covariant and contravariant vector fields. It seems that contravariant fields are just the normal vector fields they introduced in multivariable calculus, but if so, I can't figure out what covariant fields are.

I can't give u the real explanation behind it all,because i don't know how much of Diff.Geom.u know and there could be the danger of u not understanding it.My question is:are u familiar with manifolds??If so,then u'd understand it.

Daniel.

 

[pmb_phy]

I'm starting to learn differential geometetry on my own, but I'm having a little trouble figuring out the difference between covariant and contravariant vector fields. It seems that contravariant fields are just the normal vector fields they introduced in multivariable calculus, but if so, I can't figure out what covariant fields are.

See - http://www.geocities.com/physics_world/ma/intro_tensor.htm

Pete

 

[mathwonk]

in the expression f(x) you can vary either x or f. one of these varies covariantly and the other varies contravariantly. if you hunt around this site you wil find a large number of posts on this topic.

actually if you are in the habit of taking dot products for granted, it is hard to tell the difference between covariant and contravariant fields.

if v is an ordinary vector (contravariant?) then the expression v.( ) is a covariant vector. i.e. you have changed the role of v, from x to f, by introducing the dot product. as a vector v is something you can act on by a functional, but in the expression v.( ), you have set up v to act on another vector which goes in the parentheses.

now in abstract differential geometry, one does not always assume a dot product is given, so then you have to think of vectors and covectors as different objects. even if you do have a dot product, v and v.( ) should be understood as different objects. i.e. one of them is a vector and the other is a map from vectors to numbers.

subject to correction by pete, i will try to relate this to the indicial version on his site. a dot product is a "metric" and has representation of form gij, where ij are subscripts. this means it is covariant of order 2. a contravariant vector v (of order one) has local expression of form v^i, where here i is a superscript. then putting the two together we represent the covector v.( ) as:
summation v^i gij = aj, where we have summed over i. this object has one unsummed index j as a subscript, whereas v had one index as a superscript.

this difference in location of the index, tells us the difference between the contravariant vector v, and the covariant vector v.( ).

or as hurkyl sugests, think of vectors as columns, and think of functionals which act on vectors as rows. in this version, if v is a column vector, then v.( ) is merely the transpose of that column vector.

 

[Hurkyl]

One rather cheesy way of looking at it is that contravariant vectors are column vectors, and covariant vectors are row vectors. (Or do I have that backwards?) Of course, this doesn't work in full generality, but IMHO it's a good start at separating the two notions... although it does force you to carefully determine when which is appropriate. (e.g. the gradient of a scalar function is a row vector)

 

[gazzo]

covariant = row vector; because COvariant sounds like ROWvariant.
contravariant are left to columns.

hehe :) someone said that in a previous post and it's stuck. thumbs up for passive studying on physicsforums!

 

[marlon]

Covariant also means that the properties of some "object" (like a vector) on some manifold will remain the same under certain transformations that position this object on several locations on the manifold. For example when a manifold is curved, you will always need to define some space time continuum locally on that manifold. Then you need to know that the connection between these two points on the manifold is covariant so you can be certain that for example a 0-vector in one point of this manifold will still be a 0-vector in the other point.

Or, for example, look at tensors, which are a generalization of mathematical objects like vectors, matrices,...So a tensor is basically any object with indices (like a vector or matrix or double product of two or more vectors) AND certain transformation properties. Basically, these properties are that a tensor is covariant when it is transformed. Thus, this means that for example when the elements in a tensor (like the row and colomn-elements in a matrix) are 0 in one point on the manifold, they must remain zero at some other point on the manifold.

The most famous examples are the Riemanntensor and the Riccitensor from General Relativity. The Riemanntensor can be used in order to eveluate whether a manifold is curved or not. You can use it to check that the Earth is curved although in your local reference frame, everything seems flat (you can see as far as the horizon).

This tensor is calculated by using the Christoffel-symbols which are also objects with indices BUT no tensors because they are not covariant. Their structure will change as you move them along the manifold.


regards
marlon

 

[StatusX]

I can't give u the real explanation behind it all,because i don't know how much of Diff.Geom.u know and there could be the danger of u not understanding it.My question is:are u familiar with manifolds??If so,then u'd understand it.

Actually, I'm not sure. I mean, I could give you plenty of examples of manifolds, but I'm not exactly sure what isn't a manifold. I was told that the definition of a manifold was that it could be covered with open sets, and each of these sets could be related by a smooth, one-to-one function to some subset of E_n. First of all, do I have this right?

I think this excludes objects with kinks. But what about multiple objects, like two open balls separated by some distance? Or a closed ball, which can be covered with open sets relative to itself, but not relative to the ambient space? Or a cylinder or cube (as surfaces (I don't think so) or as volumes (I think so)), with kinks on the edges? Or a sheet that intersects itself? (don't think so)

Also, I heard that for two manifolds with different coordinate functions to have "the same smooth structure," there must be a smooth invertible function between the different coordinate systems (and also, ther must be one for any points where two local coordinate patches overlap on a single manifold). So for example E₁ has different smooth structure with x as its function than with x³. What does this mean? Isn't E₁ always just E₁?

Getting back to covariant vs contravariant, I will read a little more (thanks for that site by the way Pete) and get back with more specific questions. Thanks for your help so far.

 

[jcsd]

Covariant also means that the properties of some "object" (like a vector) on some manifold will remain the same under certain transformations that position this object on several locations on the manifold. For example when a manifold is curved, you will always need to define some space time continuum locally on that manifold. Then you need to know that the connection between these two points on the manifold is covariant so you can be certain that for example a 0-vector in one point of this manifold will still be a 0-vector in the other point.

Or, for example, look at tensors, which are a generalization of mathematical objects like vectors, matrices,...So a tensor is basically any object with indices (like a vector or matrix or double product of two or more vectors) AND certain transformation properties. Basically, these properties are that a tensor is covariant when it is transformed. Thus, this means that for example when the elements in a tensor (like the row and colomn-elements in a matrix) are 0 in one point on the manifold, they must remain zero at some other point on the manifold.

The most famous examples are the Riemanntensor and the Riccitensor from General Relativity. The Riemanntensor can be used in order to eveluate whether a manifold is curved or not. You can use it to check that the Earth is curved although in your local reference frame, everything seems flat (you can see as far as the horizon).

This tensor is calculated by using the Christoffel-symbols which are also objects with indices BUT no tensors because they are not covariant. Their structure will change as you move them along the manifold.


regards
marlon

Marlon, that's right, but it's also not quite right!

A covariant vector is specifically a vector which transforms with the basis vectors, a contravariant vector on the other hand is a vector that transforms against the basis vectors. If we talk about something like Lorentz covariancy, the meaning is sligthly different. Just to illustarte how confusing this can be: a contravariant Lorentz vector is not a covariant Lorentz vector though it is a vector that is Lorentz covariant!

 

[dextercioby]

Marlon, that's right, but it's also not quite right!

A covariant vector is specifically a vector which transforms with the basis vectors, a contravariant vector on the other hand is a vector that transforms against the basis vectors. If we talk about something like Lorentz covariancy, the meaning is sligthly different. Just to illustarte how confusing this can be: a contravariant Lorentz vector is not a covariant Lorentz vector though it is a vector that is Lorentz covariant!

Okay,people,let's leave the physics aside and give the definitions:
A contravariant vector in a point "p" of a manifold (a.k.a. 'vector') is an element of the vector space tangent to the manifold in the point "p":T_{p}.It can be looked upon as a 'hyperplane',due to the simple case when its dimension (and implicitely the manifold's dimension) is 2.
A covariant vector in a point "p" of a manifold (a.k.a. "one form") is an element of another vector space,called cotangent space in the point "p" of the manifold:T^{*}_{p}.Unfortunetely it does not have a geometrical representation like the tangent space.That is we cannot visualize the cotangent space not even for the simple case when 'd=1'.
Since those 2 are vector spaces we can define bases and formulate how these 'animals' behave to a change of basis.

'Lorentz covariance' is just a confusing syntagma.One who knows Diff.Geom. will never confuse with the real meaning of "covariance".I hope... Anyway,i'm sure Marlon has everything clear.

Daniel.

 

[jcsd]

And the tangent and cotangent spaces at a single point are dual vector spaces. To me that's probably the easiest way to get a basic grasp on the difference.

I think though that the catergory theorists would get pretty nasty if it was suggested that the differential geometry defintinn of covaraint and contravariant was the real deifinition.

 

[marlon]

Marlon, that's right, but it's also not quite right!

A covariant vector is specifically a vector which transforms with the basis vectors, a contravariant vector on the other hand is a vector that transforms against the basis vectors. If we talk about something like Lorentz covariancy, the meaning is sligthly different. Just to illustarte how confusing this can be: a contravariant Lorentz vector is not a covariant Lorentz vector though it is a vector that is Lorentz covariant!

Correct jcsd, i admit there is confusion possible here and that's why your point is completely correct here...though when i was referring to covariance i did not make any reference to Lorentz-covariance which is ofcourse a very important issue in GTR. But it is not fundamental since it is defined purely based upon the notion of covariance...at least that is the way i have learned in college and i cannot say that no other explanation or introduction of this concept is possible...This is just the way i keep all these concepts on the right track...

regards
marlon

ps : indeed : the notion of "transforming along with basic vectors is a very good point and this was what i wanted to state in my post"

 

[pmb_phy]

I'm starting to learn differential geometetry on my own, but I'm having a little trouble figuring out the difference between covariant and contravariant vector fields. It seems that contravariant fields are just the normal vector fields they introduced in multivariable calculus, but if so, I can't figure out what covariant fields are.

There are several ways to define these quantities. A "contravariant vector" is what you'd simply call a "vector" which comes in two flavors. There are displacement vectors, used mostly in flat spaces such as the flat spacetime of SR and then there are tangent vectors which one uses in curved spaces such as the curved spaces one often finds in GR.

A covariant vector is then defined as a multilinear operator which maps contravariant vectors into scalars. In this sense a a covariant vector is a vector in the abstract mathematical sense of the term.

Pete

 

[jcsd]

Correct jcsd, i admit there is confusion possible here and that's why your point is completely correct here...though when i was referring to covariance i did not make any reference to Lorentz-covariance which is ofcourse a very important issue in GTR. But it is not fundamental since it is defined purely based upon the notion of covariance...at least that is the way i have learned in college and i cannot say that no other explanation or introduction of this concept is possible...This is just the way i keep all these concepts on the right track...

regards
marlon

ps : indeed : the notion of "transforming along with basic vectors is a very good point and this was what i wanted to state in my post"

Yes Lorentz covariance is a bit of a red herring, I used it becasue of the unfortunate terminology of covariant Lorentz vectors and thee fact taht all Lorentz vectors are Lorentz covariant where in each it is infact a different property that is being described as 'covariant'. Really I should of talked about general covaraincy as a Lorentz tensor is not necessarily a true tensor.

What confused me is that you seemed to suggets that all tensors are covariant which is true in the sense that a general tensor is general covariant, but in the context of covaraince as used in the original question is not true.

At the moment I'm trying to wrap my head round the basics of catergory theory and to muddy the water even further (though I must admit I've only got as far as the basic definition of a catergory) it appears that it catergory theory what are callled contravariant vectors would be described as covaraint objects and what are called covaraint vectors are infact contravariant objects (I'm afraid though you'll have to ask mathwonk why that is)!

 

[jcsd]

A covariant vector is then defined as a multilinear operator which maps contravariant vectors into scalars. In this sense a a covariant vector is a vector in the abstract mathematical sense of the term.

Pete

Yes that's one way to think of them and it's the most formal defintion that I know of these objects (though it's best to call covaraint vectors linear functionals as multilinear usually implies more than one variable and the term oeprator usually implies a map from vector space to vector space)

 

[gvk]

Okay,people,let's leave the physics aside and give the definitions:
A contravariant vector in a point "p" of a manifold (a.k.a. 'vector') is an element of the vector space tangent to the manifold in the point "p":T_{p}.It can be looked upon as a 'hyperplane',due to the simple case when its dimension (and implicitely the manifold's dimension) is 2.
A covariant vector in a point "p" of a manifold (a.k.a. "one form") is an element of another vector space,called cotangent space in the point "p" of the manifold:T^{*}_{p}.Unfortunetely it does not have a geometrical representation like the tangent space.That is we cannot visualize the cotangent space not even for the simple case when 'd=1'.
Since those 2 are vector spaces we can define bases and formulate how these 'animals' behave to a change of basis.
Daniel.

About geometrical representation.
If you make a sketch of tangent space for manifold, you do simultaneously the sketch of cotangent space, because the tangent vector looks exactly the same as cotangent covector, e.g. for 'd=1' tangent line coinside with cotangent line, for 'd=2' tangent plane coinside with cotangent plane. Note, that this is true if you have a metric in the vector space. If you don't have a metric, strictly speaking, you can not draw tangent space.
But you can pretend that you can do it even you don't have a metric. In this case, the best way to cheat your imagination is to draw the second line, plane, hyperplane,... parallel to the tangent space (line, plane, hyperplane,...). It's not the best way, of course, but, at least, it gives a sense that those spaces are different and connected.
This 'geometrical representation' is used by many authors.
Many people emphasized that tangent and cotangent spaces are different. It is true, but, in my view, it is equally important to emphazise, that those 'animals' are tightly connected. If you change somehow the basis in tangent space, the basis of cotangent space is simultaneously changed. And these changes go in exactly oposite direction, so the scalar product of any vector from tangent space and any covector of cotangent space does not depend on the basis. And these values are called invariants. The scalar product between two vectors from different spaces (tangent and cotangent) looks very strange, if even the scalar product inside of each space is prohibited.
But it looks normal if one remember a tight connection between them.
In fact, the introduction of tangent and cotangent spaces allows to bypass the metric, which was the main tool to get the 'physical' invariants.

 

[mathwonk]

asusual everyone "visualizes" or imagines even precise mathematical objects entirely differently. i have trouble even understanding the comments here on the ease or impossibility of visualizing tangent and cotangent vectors. here is my personal view of them.

i imagine say a two dimensional tangent space as a plane, with a distinguished point, the origin. then a non zero cotangent vector, being by definition a non zero linear function on this space with real values, is determined up to a constant multiple by the subspace of tangent vectors which are mapped to zero, hence by a line through the origin. so the projective cotangent space is merely the set of lines through the origin of the tangent space. to determine the linear function fully and hence the covector, we need to know which vectors are mapped to 1, which forms a line parallel to the previously given line through the origin. so in this representation, a non zero covector is merely a line in the plane not passing throuigh the origin.

in higher dimensions, it is a hyperplane not passing through the origin. in case one has a notion of perpendicularity, one can draw a line through the origin perpendicular to this hypwerplane, and then identify the hyperplane with the intersection point of line and hyperplane. this allows one to view the tangent space and cotangent space as the same, not otherwise.

now my grandaughter wants to, play pbskids. so goodbye

 

[gvk]

In my understanding, 'to visualize' (in a simplest significance) means to form a mental geometrical image, which can be ploted on the paper or builded as a 3-D model.
Let's consider again how we 'visualize' the cotangent space. It is a vector space having the basis, but this basis is builded by using the basis of direct vector space. For 1-d it is e^1(e_1)=1, where e_1 is basis vector of tangent space, e^1() means a linear functional. Now we plot the straight line with origin and vector e_1 comming out from origin. It lies on the line and this line is tangent space. Now we try to plot on the same plane the cotangent space. If we do this we automaticaly introduce the metric relations between tangent and cotangent spaces (scalar product!), because we care about orientations between two lines, at least. So, we write the linear functional as the scalar product (e^1, e_1)=1. If the lines are perpendicular, the scalar product is 0 and e^1 is not basis of cotangent space. It's not what we wanted, but if the lines are coinsided, it is OK. Now we take another the basis in tangent space e'_1=\lambda e_1. The basis in cotangent spaces will be changed, e'^1=\frac{e^1}{ \lambda}, otherwise we can not satisfy the condition (e'^1, e'_1)=1.
The similar picture can be in 2-d case, where (e^1, e_1)=1 , (e^2, e_2)=1, (e^1, e_2)=0, (e^2, e_1)=0. But of course, it does not mean that pair e^1 and e_1 (or e^2 and e_2) should be parallel. Here e^1, e^2 constitute the reciprocal basis in the same plane.

 

[StatusX]

I have a question. I'm sorry if this is a little off topic, but I didn't think it deserved it's own thread. The metric is defined as:

g_{ij} = \hat e_i \cdot \hat e_j

where the \hat e_i are the basis vectors of the local coordinate system. In terms of the ambient coordinates, this is:

(\hat e_i)_s = \frac{\partial y_s}{\partial x^i}

But this is where I get confused, because sometimes the metric is given, such as in a minkowski space where its diag[1,1,1,-c^2]. But the paper I'm reading says that:

g_{ij} = \hat e_i \cdot \hat e_j = \frac{\partial y_s}{\partial x^i} \frac{\partial y_s}{\partial x^j}

But this is implies the ordinary dot product is being used, and it can only have the normal signature (1,1,1,1). What am I missing?

 

[jcsd]

It is the dot product, in this case the dot product of the time basis vector and itself in the basis you have choosen is -c^2 and it is the metric that defines this.

 

[StatusX]

so are you saying the last equal sign in my post isn't always true? When is it? is it necessary and sufficient that the signature be (1,1,1,...,1)?

 

[mathwonk]

gvk, to me you are missing the point that choosing a basis is a very unnatural operation, which then also chooses an unnatural isomorphism between tangent and cotangent spaces. this causes me to ask: how much can you discuss about this topic without introducing bases? that would annul pretty much your entire last post. i suspect that overuse of bases is a prime rason some people here confuse tangent and cotangent spaces, and other topics in linear algebra.

 

[mathwonk]

to emphasize: once you choose a basis, any (real) vector space is then canonically isomorphic to R^n. Of course it IS true that R^n is naturally isomorphic to its dual, and there is a canonical dot product in R^n, but these things are not at all true in vector spaces in general.

the same problem occurs for differential manifolds. although local coordinates exist, there is no NATURAL choice of local coordinates in a differential manifold, for example on a sphere, or even a circle, there is no natural choice of origin. and although riemannian mettrics exist on any (paracompact) manifold, there is also no NATURAL choice of riemannian metric, so again the only natural intrinsic phenomena are those that can be discussed without choosing local coordinates.

so i conjecture that the reason many people fail to appreciate the difference between concepts like covariance and contravariance is they assume the existence of local coordinates, and bases. a physicist should appreciate this more than anyone, since local coordinate systems do not occur in nature, they are imposed for our convenience.

 

[saski]

Let me tell a story. Its relevance will become apparent.

When I had my first taste of analytic geometry, I got the idea that if

(x,y)

were the coordinates of a point, and

a x + b y = 0

was the equation of a line, then

(a,b)

should be "the coordinates of a line".

But there were problems. For one thing, (k a, k b) represented the same line (for k != 0). For another, (a,b) = (0,0) wasn't a line, but the whole plane. Thirdly, there were all these other lines, a x + b y + c = 0, so perhaps the line coordinates really should be (a,b,c); but there was no meaningful corresponding set of "enhanced points", (x,y,h).

I can smile at myself now, but in fact I had a grasp of a couple of important principles. Such as isomorphism: the whole set of coordinate tuples had to be in one-to-one correspondence with the things they represented. Also, the duality implied by a formal relabelling of variables, i.e. exchanging (a,b) with (x,y).

I didn't have the vocabulary or the discipline at the time; but if I'd perservered, there are three ways I could have gone.

(1) If what I really wanted was to represent the Euclidean plane, I would have accepted that the points were isomorphic to R^2, while the lines were isomorphic to R^3 modulo scalar multiples; that's just how things are.

(2) If I had insisted on my "symmetry" between points and lines, I might have discovered the representation of points by homogeneous tuples (x,y,h). Then, reworking all the formulas, I would have found the true duality of points and lines (or generally hyperplanes) in projective space.

(3) Or I could have insisted on identifying the class of geometric objects to which all (a,b) in R^2 were isomorphic. The answer of course is that (a,b) corresponds to the function:

(x,y) \mapsto a x + b y

Euclid had no name for such functions. Euclidean geometry comprises just ruler-and-compass constructions; and a high school analytic geometry unit is concerned with demonstrating the power of equations to model the same points, lines and curves. It's no wonder that an entity which covers the whole plane should have been left out. The graph of the function, if we throw in an extra coordinate for its value, is an inclined plane, having constant slope. Just for the momemt, let us call such a function a "slope".

My original problems are resolved. The line a x + b y = 0 is not the function (x,y) \mapsto a x + b y; it's the kernel of that function, and the different function (x,y) \mapsto k a x + k b y has the same kernel. (x,y) \mapsto 0 x + 0 y is a function of the same kind, its kernel being the whole plane. And we can also define functions (x,y) \mapsto a x + b y + c.

All this should have been my tipoff, that there were two kinds of vectors in a cartesian coordinate system: contravariant vectors (x,y) and covariant vectors (a,b). Or column vectors and row vectors if you prefer. If I'd perservered, I would have found that you can add slopes and scalar-multiply them, in fact do all the vector tricks. I would also have realized that the cartesian coordinate system consists of two slopes,

(1,0) : (x,y) \mapsto x
(0,1) : (x,y) \mapsto y

These "projection" or coordinate-taking functions are the canonical unit slopes or unit covariant vectors.

That's the story. So we have a geometric picture distinguishing the two kinds of vectors. We don't have to develop the theory of manifolds to get it; vector spaces alone are sufficient. However we do need to take account of (1) change of basis, and (2) metric functions, to develop the full relationship between the two kinds.

As mathwonk was just saying, we are prone to traps if we start by assuming s basis, which of course is exactly what I've done. My view is that we initially have trouble seeing the difference between covariant and contravariant vectors because "classical", or "Gibbs notation" vector algebra (with unit vectors i,j,k in Clarendon bold type) teaches us first about the contravariants, and then teaches us to use them in the covariant role, for instance when we write a \bullet x = 0 as the equation for a plane in 3D. It works because we have a built-in definition of "perpendicular", so that we can use the symbol a interchangably for a slope and for a vector normal to its level surfaces. But when we get to curvilinear coordinates or differential structures on manifolds, we have a relearning exercise.

Continued on next rock.

 

[mathwonk]

Nice story.

here is another try at why covariant and contravraiant are different. Logically, covariant means in the "same direction as", while contravariant means in the "opposite direction from". Thus there is no way they can be the same. They are by definition opposites, in the sense of transforming in opposite directions.

here is the simplest illustration: consider x(t) as a function of t. Thus given a t, we can transform it into an x, i.e. x(t). But we do not therefore transform a FUNCTION of t into a function of x. just the opposite, we transform a function of x, such as f(x) into a function of t, namely into f(x(t)).

Thus the points, i.e. the coordinate variables, go from t to x, while the functions acting on the points, go the opposite way, from f(x) to f(x(t)). Thus functions and points, or functions and coordinates, transform in opposite directions.

This means if the "standard" direction is considered as the direction the coordinates go in, i.e. from t to x, then the other transformation, from f(x) to f(x(t)), should be called contravariant.

This is reflected exactly in the distinction between tangent vectors and cotangent vectors. A tangent vector at p is represented by a curve passing through p. Then if f is a mapping taking p to q, we can apply it to the curve, obtaining a curve through q. This action of f on curves, is the "derivative" of f. So the derivative of f goes in the same direction as does f, i.e. tangent vectors v at p go to tangent vectors Dfp(v) at f(p).

On the other hand, dual vectors go the opposite way. Now this is going to get more complicated notationally, and I apologize. But here goes:

For instance, even if we use an inner product to represent a cotangent vector at f(p) as dotting with a tangent vector w at f(p), i.e. say we think of <w, > as a cotangent vector at f(p), it still transforms the other way, i.e. from the q's back to the p's.

I.e. given w = a tangent vector at f(p), if we use the dot product to consider it as the cotangent vector <w, > at f(p), then it gives us a cotangent vector Dfp*(<w, >) at p as follows: to prove Dfp*(<w, >) is a cotangent vector at p, we have to show how it acts on a tangent vector v at p.

Well, given any tangent vector v at p, the pullback covector Dfp*(<w, >) acts on v by first mapping v over to the tangent vector Dfp(v) at f(p), and then applying <w, > to that vector.

I.e. Dfp*(<w, >)(v) = by definition, <w, Dfp(v)>. Thus DOTTING with a tangent vector, transforms in the opposite direction from the tangent vector itself.

Now it is true that this operation on tangent vectors v at p, CAN be achieved by dotting them with some tangent vector at p, but there is NO natural choice of such! The choice depends on the choice of inner product at p, which is completely arbitrary.

I.e. it is not true that the covector Dfp*(<w, >) at p, obtained by pulling back <w, >, is in any natural way equal to dotting with a tangent vector at p. on the other hand without any choice of inner product, the operation of composing the derivative of f with a linear function at f(p) is totally natural.


Oh I guess I went overboard here. but heck, it is hard to just give up. Soon school will start again and I will have no such time on my hands. I wil be engaged trying to convince people that there is no one distinguished "dependent vector" in a dependent set of vectors.

Actually it is the same idea, since my whole point is that covariant and contravariant are not properties of a single type of vector, but of a relationship between two things. I.e. to detect covariance you have to compare transformation rules of your object, with those of a standard object, usually the coordinate map on points.


Of course classical differential geometry terminology has screwed this whole covariant contravariant thing up BIG time, and uses the terms backwards. i.e. in differential geometry, "contravariant vectors" are the tangent vectors that transform in the SAME direction as the mapping on points, while "covariant vectors" or "covectors" are the ones that transform in the opposite direction. I.e. in classical differential geometry language, "contravariant vectors" transform covariantly, because Dfp goes in the same direction as f, while "covariant vectors" transform contravariantly, since Dfp* goes in the opposite direction from f.

Of course algebraic topologists are also guilty since "cohomology" is a contravariant operation. Years ago Peter Hilton tried to change history and call it "contrahomology", but the reason you have never heard of contrahomology, is of course he failed.

No matter, it still follows that covariant and contravariant vectors are distinct because they transform in the opposite direction from EACH OTHER.

(Maybe the classical screwup occurred because classicists were not in possession of the idea of coordinates transformations as maps on points, and were instead referring to the transformation of coordinate FUNCTIONS as opposed to the points of coordinate space. So they were being consistent, in calling contravariant vectors ones which transformed in the opposite direction to the coordinate functions. So possibly the whole confusion began, and persists, by substituting notation, i.e. coordinates, in place of concepts, i.e. geometry.)

 

[StatusX]

I'd really appreciate it if someone could address my last post, I'm still very confused. Thanks.

 

[gvk]

I have a question. I'm sorry if this is a little off topic, but I didn't think it deserved it's own thread. The metric is defined as:

g_{ij} = \hat e_i \cdot \hat e_j

where the \hat e_i are the basis vectors of the local coordinate system. In terms of the ambient coordinates, this is:

(\hat e_i)_s = \frac{\partial y_s}{\partial x^i}

But this is where I get confused, because sometimes the metric is given, such as in a minkowski space where its diag[1,1,1,-c^2]. But the paper I'm reading says that:

g_{ij} = \hat e_i \cdot \hat e_j = \frac{\partial y_s}{\partial x^i} \frac{\partial y_s}{\partial x^j}

But this is implies the ordinary dot product is being used, and it can only have the normal signature (1,1,1,1). What am I missing?

You are considering other space than Euclidean with the metric (1,-1,-1,-1,). The space in your case is called pseudoeuclidean. Its 1st quadratic form is not possitively defined.

 

[gvk]

Any compact manifold embedded in Euclidean space can have Riemannian metric with positively defined 1-st quadratic form. And any compact manifold with Riemannian metric can be a submanifold of a Euclidean space (it is so called Whitney theorem).
If the 1st quadratic form is not possitively defined it can not be embedded in Euclidean space with the metric (1,1,1,1) (see e.g. very nice book L.P. Eisenhart Riemannian Geometry).

 

[gvk]

Mathmonk,
Yes, the choosing a basis is "a very unnatural operation", because merely there is no any preferences between two different coordinate systems. This is why the notion of "the transformation of coordinate system" is appeared in mathematics. Unfortunatly, we can not ignore the coordinate system method and this would be the full answer to your question.

 

[gvk]

StatusX
To understand the metric of embedded manifolds yourself (it looks like you are thoughtful man, because you give very good questions) take any ambient pseudo-Euclidean space flat space and write the square of infinitesimal distance:

(ds)^2 = \sum_{i=0}^{n} \delta_i (dx^i)^2

where the \delta_i are the signature (+-1). Then take any embedded manifold:

x^i = x^i(q_1, q_2,...q_m), m &lt; n

and substitute their differentials in the infinitesimal distance. You will get the expression for the local metric, which is different from yours

g_{ij} = \sum_{k=0}^{n} \delta_k \frac{\partial x^k}{\partial q^i} \frac{\partial x^k}{\partial q^j}

But you can write it exactly in your form, if for negative signature you introduce the imaginary coordinate

y^k = i * x^k

and think that all signatures are positive. So, this may be confused you.
It is interested to note that embedded manifolds of the flat space with indefinite 1-st quadratic form can have the Riemannian metric. But as I mentioned before, the vice versa statement is wrong.

 

[StatusX]

gvk, thanks for your replies. I'm only just beginning, so I don't understand all of what you're saying (quadratic forms, for example). But what you said about imaginary numbers seems to make sense. The only thing, though, is that the source I'm reading doesn't mention imaginary numbers. What it does do is define the first christoffel symbol in these two, allegedly equivalent ways:

[p q, r] = \frac{1}{2}\left(g_{qr,p} + g_{rp,q} - g_{pq,r}\right)

[p q, r] = y_{s,pq} \ y_{s,r}

where, eg.,
y_{s,r}=\frac{\partial y_s}{\partial x_r}

Is that second definition always valid?

I guess my central misunderstanding is about what the ambient space is when the metric isn't positive definite. Is it still euclidean space? Does it have this strange metric as well? Or is the metric just a property of a specific manifold?

 

[mathwonk]

gentlemen,

i have made this subject as clear as I can ever do.

best wishes,

mathwonk. but mathmonk might work too.

 

[gvk]

StatusX,
Thank you for these questions!
You just got to the central point of the Differential Geometry.
I guess you are familiar with basics of field theory and quantum mechanics.
If you do, the notion of 'the space without metric' does not shock you. If you don't you need to read some books of this subject to feel more comfortable yourself.
Yes, the second formula of Christoffel symbol is valid for any space (with any metric or without any metric). Why? Because it can be received from
simple requirement that result of differentiation of the vector or covector should be a tensor. It is not normal partial differentiation, it's called 'covariant differentiation'.
It turned out that the covariant differentiation has deep and profound meaning. First discovered by Ricci in 90th of 19 century but only in 30th with the development of quantum mechanics it was understood that this notion is deeper than the metric properties. And this can be the starting point of new approach to diff.geometry which many people try to pursue. (See e.g. the posts of mathmonk)

However, I'll try to explain this notion in simple words. Let we have a vector or covector in one point of space. Assume the space has some (e.g. curvilinear) coordinates. The purpose of Ricci was to receive the vector or covector in another point which is close to the origin one. Oh, yes, for this we need to find differential. But the normal differentiation does not work because the coordinate lines are curved and normal partial derivitive destroys your vector, it won't be the vector anymore (you can check it youself). To create the vector in the neighboring point Ricci turned slightly the vector (or covector) toward the curved line in order the angle between vector and coordinate line be the same. This new vector or covector becomes the real vector or covector. They satisfy the transformation properties. This procedure is called 'parallel transport'. Of course, in eucleadian space where the coordinate systems are stright lines, the covariant differentiation is exactly the same as normal differentiation. The difference between the normal differential and 'covariant differentiation' is defined by Christoffel symbols.
Because 'covariant differentiation' was invented to connect vector in one point with vector in another point, it's called sometimes 'connection'.
I emphasize that the connection does not require any metric properties, it can be received from the following relations:
A) the operation of covariant differentiation is linear,
B) the result of covariant differentiation of the tensor forms a tensor again.

 

[Peterdevis]

Because 'covariant differentiation' was invented to connect vector in one point with vector in another point, it's called sometimes 'connection'.
I emphasize that the connection does not require any metric properties, it can be received from the following relations:
A) the operation of covariant differentiation is linear,
B) the result of covariant differentiation of the tensor forms a tensor again.

This is correct. So you can define a large number of connections on a manifold and each of them implies a distinct notion of covariant differentiation.
When there is a metric you can define one unique connection that is torsion free and metric compatibele:

[p q, r] = \frac{1}{2}\left(g_{qr,p} + g_{rp,q} - g_{pq,r}\right)



Thats the one used in GR

 

[saski]

Hiya StatusX et all

The two expressions for [pq,r] are equivalent, and this is independent of the metric signature.

I had a little trouble interpreting the y-coordinate notation, but I think it makes sense. Assuming so, the formal proof goes as follows.

We start with a set of basis vectors e_i, having components in the global coordinate system:

{e_i}_s = \partial y_s / \partial x_i = y_{s,i}

so that

g_{ij} = {e_i}_s \bullet {e_j}_s

where we using the summation convection over s. Now if we are simply summing over s, that means we're assuming a metric signatire that's all 1's. But we can easily change that assumption later.

So we have

g_{pq} = y_{s,p} y_{s,q}

We can then use the product rule to write:

g_{pq,r} = y_{s,pr} y_{s,q} + y_{s,p} y_{s,qr}

g_{qr,p} = y_{s,qp} y_{s,r} + y_{s,q} y_{s,rp}

g_{rp,q} = y_{s,rq} y_{s,p} + y_{s,r} y_{s,pq}

For derivatives with respect to coordinates, the second derivatives commute: {...}_{,ij} = {...}_{,ji} .

So we rewrite the above as:


g_{pq,r} = y_{s,pr} y_{s,q} + y_{s,p} y_{s,qr}

g_{qr,p} = y_{s,pq} y_{s,r} + y_{s,q} y_{s,pr}

g_{rp,q} = y_{s,qr} y_{s,p} + y_{s,r} y_{s,pq}

Notice how the first term in the first equation matches the second term in the second, and likewise the second in the first matches the first in the third, and the first in the second matches the second in the third.

If we define

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}}

then terms cancel so that we get:

[pq,r] = y_{s,pq} y_{s,r}

as required.

This derivation works because the original expression for g_{pq} is symmetric in p and q. If we were treating a non-symmetric bilinear function

h_{pq} = A_{st} y_{s,p} y_{t,q}

with a constant A, we wouldn't be able to match terms like that.

Now with a metric whose signature is not {1,1,...}, but say {-1,1,...}, we would have to write it as

g_{pq} = \eta^{st} y_{s,p} y_{t,q}

where \eta^{11} = -1, \eta^{22} = 1, etc, and \eta^{st} = 0 where s != t.

But we can thread the \eta term through the derivation, making each term

\eta^{st} y_{s,-} y_{t,-}

and still match the terms, because \eta is symmetric in s and t.

So, to express [pq,r] for metrics of indefinite signature, one should really write

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}} = \eta^{st} y_{s,pq} y_{t,r}

 

[StatusX]

Now with a metric whose signature is not {1,1,...}, but say {-1,1,...}, we would have to write it as

g_{pq} = \eta^{st} y_{s,p} y_{t,q}

where \eta^{11} = -1, \eta^{22} = 1, etc, and \eta^{st} = 0 where s != t.

But we can thread the \eta term through the derivation, making each term

\eta^{st} y_{s,-} y_{t,-}

and still match the terms, because \eta is symmetric in s and t.

So, to express [pq,r] for metrics of indefinite signature, one should really write

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}} = \eta^{st} y_{s,pq} y_{t,r}

Thank you saski, this is what I've been looking for all along. the eta term was never mentioned in my source, that was the source of my confusion.

 

[gvk]

Hiya StatusX et all

The two expressions for [pq,r] are equivalent, and this is independent of the metric signature.

....

Now with a metric whose signature is not {1,1,...}, but say {-1,1,...}, we would have to write it as

g_{pq} = \eta^{st} y_{s,p} y_{t,q}

where \eta^{11} = -1, \eta^{22} = 1, etc, and \eta^{st} = 0 where s != t.

But we can thread the \eta term through the derivation, making each term

\eta^{st} y_{s,-} y_{t,-}

and still match the terms, because \eta is symmetric in s and t.

So, to express [pq,r] for metrics of indefinite signature, one should really write

[pq,r] = (1/2) { g_{qr,p} + g_{rp,q} - g_{pq,r}} = \eta^{st} y_{s,pq} y_{t,r}

Saski,
No, two expressions are not equivalent. First is dependent on the metric. It is 'specific'. The second is not. It is valid for any connection, even without any metric. They will be equivalent if you take the 'specific' connection which called COMPATIBLE with metric (see the post of Peterdevis).
To be compatible with metric the connection \triangledown (covarient differentiation) should satisfy the condition:

\triangledown_r g_{pq} =0.

The first expression

[pq,r] = (1/2) ( g_{qr,p} + g_{rp,q} - g_{pq,r})

is compatible with metric, and include the signature inside the notation g_{pq}. The metric can be pseudo-Riemannian (not definite), so you don't need to modify the formula to calculate the compatible connection.

 

[saski]

Thanks gvk, I'm alerted. I'll need some time to think this over, before I post back. It's a very important issue.

 

[StatusX]

So let me see if I have this straight. It is always possible to define a christoffel symbol like this:

[pq,r] = y_{s,pq} y_{s,r}

Even if you don't have a metric. But if you do have one, you can alternatively use this:

[pq,r] = \frac{1}{2} (g_{qr,p} + g_{rp,q} - g_{pq,r})

or equivalently:

[pq,r] = \eta^{st} y_{s,pq} y_{t,r}

In GR, they use the second definition. Is this right?

 

[gvk]

StatusX,
What kind of source (book) do you follow in your study of DiffGeom? You don't need the 3rd formula, which contains some incorrection. The second one is perfectly good for caclulation of connection's coefficients in space with indefinite metric. It is OK for GR.
If you don't have some restictions in using the source I can recommend you some good book (of course, from mine point of view).

 

[gvk]

StatusX, Saski
Add one thing about general connection which seems to be important.
If you want to receive the connection from the relations (not involving a metric):
you will get acctualy the Christoffel symbol of the second kind:

\Gamma_{ij}^{k} = \frac{\partial^2 y^m}{\partial x^i \partial x^j} \frac{\partial x^k}{\partial y^m}

You wrote the expressions for the Christoffel symbol of the first kind

\Gamma_{ij},_{k}.

The well known relation between them (rising lowing indices) exists only for connection compatible with a metric. So, you need to correct the first formula too.

 

[StatusX]

OK, I think I understand why I was confused. Either a metric or an ambient space are specified, not both. For example, you can't have a manifold embedded in euclidean space with a minkowski metric. For a general manifold which isn't defined in terms of an ambient space, the metric presents a way of "pretending" there is one. Do I finally have this right?

If so, then does the definition of the christoffel symbol involving y's (ambient coordinates) apply in non-euclidean ambient spaces? Or do you simply use the special metric and not worry about ambient coordinates, just acknowldeging the ambient space isn't euclidean. I apologize if this was so painfully obvious you assumed I realized it, buy my source (which is http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/pdfs/DiffGeom.pdf by the way, and I'm using it because it's free and I can't afford to buy a book right now) was very confusing in this respect. Thanks a lot for your help so far.

 

[gvk]

OK, I think I understand why I was confused. Either a metric or an ambient space are specified, not both.
For example, you can't have a manifold embedded in euclidean space with a minkowski metric. For a general manifold which isn't defined in terms of an ambient space, the metric presents a way of "pretending" there is one. Do I finally have this right?
If so, then does the definition of the christoffel symbol involving y's (ambient coordinates) apply in non-euclidean ambient spaces? Or do you simply use the special metric and not worry about ambient coordinates, just acknowldeging the ambient space isn't euclidean. I apologize if this was so painfully obvious you assumed I realized it, buy my source (which is http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/pdfs/DiffGeom.pdf by the way, and I'm using it because it's free and I can't afford to buy a book right now) was very confusing in this respect. Thanks a lot for your help so far.

Yes, you are almost right, exept for the case that metric is specified in addition to the ambient space (coordinate system).
I wrote the Christoffel symbol which depend only on transformation rules from y's to x's and does not involve any metric. Of course, it's far away from 'perfect' definition, but it can give you a sense that Chistoffel symbols and, hence, connection can be defined without metric. The transformation's rule of Christoffel symbol may serve as a general definition for this notion itself:

\Gamma_{ij}^{k} = \frac{\partial x^k}{\partial y^m} ( \Gamma_{ls}^{m} \frac{\partial y^l}{\partial x^i} \frac{\partial y^s}{\partial x^j}+ \frac{\partial^2 y^m}{\partial x^i \partial x^j})

It is the same kind as the definition of vector or covector through the transformation rules. And here the same relation to the controvesial 'visualization' problem. Here, the metric is the only condition to visualize the connection as a 'parallel transport'.
I think, your source gives you the level of knowledge up to 20-th. In my opinion, the best book of such type is the 'classical' book Levi-Chivita, Absolute Differential Calculus. Blackie&Son Limited London Glasgo 1927. It include all detail about metric Riemmannian and Minkowski space and GR and
written in clear 'nonformal' language.
I would also recommend the book Dubrovin, Novikov, Fomenko,
Modern Geometry-Methods and Applications, Part I,II,III (Universitext)
Springer-Verlag (1990). The part I would be enough for first reading. It is written in 'nonformal' language for physicists. All books you can borrow from a library.

 

[makris]

Even though I am posting this quite late, I have decided to post it hopping that I will save some people's time.
I have found the most lucid explanation on the covariant - contravariant issue in the book
"Tensor Analysis and Continuum Mechanics" by Wilhelm Flugge. (pages 2-7).

The explanation there assumes only the dot product of two vectors and the work (Force* distance) as prerequisites. No knowledge about manifolds or relativity is needed. If you have access to the book (via a library probably...) it will help you a lot.

 

[gvk]

Makris,
Please explain what did you learn from "Tensor Analysis and Continuum Mechanics" by Wilhelm Flugge about covariant vs. contravariant.

 

[makris]

This is a reply to the message by gvk.

Sorry I am replying this late...
The book by Wilhelm Flugge gives an introduction to tensors assuming minimum prerequisites.
It explains very well the following (pages 2-8 or a little bit more):

1.)Covariant and contravariant base vectors.
2.)Covariant and contravariant components of a vector and a second order tensor.
3.) Metric tensor.

I tried myself to read several book on tensors but there were too mathematical for me. I found the most lucid explanation on the book I mentioned. I undestood the consepts immediately despite the fact that I am neither mathematician nor physisist. It also helped me understand more advance books on the subject. From this point of view I suggested reading this part of the book. The book contains additional topics like Christoffels symbols etc but in subsequent chapters. However, I have not compared these topics with simillar ones written in other books to give my opinion.

Makris

 

[StatusX]

Just wondering, is the covariant/contravariant distinction the same as the one between bra and ket vectors in quantum mechanics?

 

[jcsd]

Yes, exactly the same distinction.

 

[Haelfix]

I always remember the distinction between the two b/c they are exactly the opposite of the logical way of assigning them.

 

[mathwonk]

Here is what i think haelfix means: if F(X) is a functor, then F is "covariant" if for any map f:X-->Y we get a map f*:F(X)-->F(Y).

On the other hand F is "contravariant" if for any map f:X-->Y we get a map f*:F(Y)-->F(X).

i.e. covariant means "same direction", and contravariant means "opposite direction".


Thus tangent spaces are covariant, since if f:X-->Y, then for each p in X we get

dfp:Tp(X)-->Tf(p)Y.

On the other hand the dual of a tangent space, i.e. the cotangent space T*(X), is contravariant, since when f:X-->Y, then we get dfp*:T(Y)*-->T(X)*.


Unfortunately this is the opposite convention from that adopted for some historical reason in diifferential geometry, and hence in physics. I.e. those subjects use the words backwards.

Does this help anyone? Sadly to me, many of the books recommended here discuss these terms purely with regard to whether the indices used to describe them are "up" or "down". Anyone wishing to become expert is advised to try to get beyond that type of mechanical understanding of the topic.

The intelligent student will not be misled by discussions which seem easier because they omit explanation of the concepts involved and discuss only the notation. Remember, one can go from an understanding of the concepts to a grasp of the notation, but not the other way.