covariant vs contravariant

**mathwonk** · Jan2-05, 04:20 PM

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** · Jan3-05, 10:48 AM

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

, where

is basis vector of tangent space,

() means a linear functional. Now we plot the straight line with origin and vector

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

. If the lines are perpendicular, the scalar product is 0 and

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 $LaTeX Code: esingle-quote_1=\\lambda e_1$ . The basis in cotangent spaces will be changed, $LaTeX Code: esingle-quote^1=\\frac{e^1}{ \\lambda}$ , otherwise we can not satisfy the condition

LaTeX Code: (esingle-quote^1, esingle-quote_1)=1

.
The similar picture can be in 2-d case, where

,

,

,

. But of course, it does not mean that pair

and

(or

and

) should be parallel. Here

constitute the reciprocal basis in the same plane.

**StatusX** · Jan3-05, 07:27 PM

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:

$LaTeX Code: g_{ij} = \\hat e_i \\cdot \\hat e_j$

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

$LaTeX Code: (\\hat e_i)_s = \\frac{\\partial y_s}{\\partial x^i}$

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:

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

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** · Jan3-05, 08:38 PM

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** · Jan3-05, 09:09 PM

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** · Jan3-05, 10:52 PM

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** · Jan4-05, 11:06 PM

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** · Jan5-05, 07:08 AM

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

were the coordinates of a point, and

was the equation of a line, then

should be "the coordinates of a line".

But there were problems. For one thing,

represented the same line (for

). For another,

wasn't a line, but the whole plane. Thirdly, there were all these other lines,

, so perhaps the line coordinates really should be

; but there was no meaningful corresponding set of "enhanced points",

.

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

with

.

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

, while the lines were isomorphic to

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

. 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

in

were isomorphic. The answer of course is that

corresponds to the function:

$LaTeX Code: (x,y) \\mapsto a x + b y$

Euclid had no name for such functions. Euclidean geometry comprises just ruler-and-compass constructions; and a highschool 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

is not the function $LaTeX Code: (x,y) \\mapsto a x + b y$ ; it's the kernel of that function, and the different function $LaTeX Code: (x,y) \\mapsto k a x + k b y$ has the same kernel. $LaTeX Code: (x,y) \\mapsto 0 x + 0 y$ is a function of the same kind, its kernel being the whole plane. And we can also define functions $LaTeX Code: (x,y) \\mapsto a x + b y + c$ .

All this should have been my tipoff, that there were two kinds of vectors in a Cartesian coordinate system: contravariant vectors

and covariant vectors

. 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 realised that the cartesian coordinate system consists of two slopes,

$LaTeX Code: (1,0) : (x,y) \\mapsto x$
$LaTeX Code: (0,1) : (x,y) \\mapsto y$

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

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 $LaTeX Code: a \\bullet x = 0$ 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

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** · Jan5-05, 03:03 PM

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** · Jan11-05, 01:11 AM

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

**gvk** · Jan11-05, 01:23 PM

Originally Posted by 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:

$LaTeX Code: g_{ij} = \\hat e_i \\cdot \\hat e_j$

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

$LaTeX Code: (\\hat e_i)_s = \\frac{\\partial y_s}{\\partial x^i}$

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:

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

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** · Jan11-05, 02:02 PM

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** · Jan11-05, 02:26 PM

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** · Jan12-05, 09:21 AM

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:

$LaTeX Code: (ds)^2 = \\sum_{i=0}^{n} \\delta_i (dx^i)^2$

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

LaTeX Code: x^i = x^i(q_1, q_2,...q_m), m < n

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

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

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

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** · Jan12-05, 12:44 PM

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:

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

$LaTeX Code: [p q, r] = y_{s,pq} \\ y_{s,r}$

where, eg.,
$LaTeX Code: y_{s,r}=\\frac{\\partial y_s}{\\partial x_r}$

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** · Jan12-05, 11:15 PM

gentlemen,

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

best wishes,

mathwonk. but mathmonk might work too.