Covariant and Contravariant Tensors

[abrowaqas]

we have studied in Tensor's analysis that there are two kinds of tensors that usually used in transformation. one is Contravariant & covariant. what is the difference between them and and why they are same for Rectangular coordinates?

 

[Fredrik]

I'll just quote myself:

Forget about manifolds for a moment and let V be an arbitrary finite-dimensional vector space over the real numbers. Now define V* as the set of all linear functions from V into $\mathbb R$. Then define the sum of two members of V*, and the product of a member of $\mathbb R$ and a member of V* by

$$(f+g)(v)=f(v)+g(v)$$

$$(kf)(v)=k(f(v))$$

These definitions give V* the structure of a vector space. It's called the dual space of V. Since V* is a vector space, the members of V* are vectors. However, when V is the tangent space of a manifold, the members of V are called "tangent vectors" and the members of V* are called "cotangent vectors". This is often shortened to the misleading "vectors" and "covectors", or worse (much worse actually) "covariant vectors" and "contravariant vectors".

See this post for comments about the definition of "tangent space".

why they are same for Rectangular coordinates?

Tangent vectors and cotangent vectors are never the same, since they live in different spaces. A cotangent vector is a function that takes a tangent vector to a number. However, if the manifold is endowed with a metric tensor, there's natural way to associate a tangent vector with each cotangent vector and vice versa. For example, if $v^\alpha$ are the components of a tangent vector v, the corresponding cotangent vector has components $v_\beta=g_{\beta\alpha}v^{\alpha}$. If the components of the metric tensor are =0 when $\alpha\neq\beta$ and =1 when $\alpha=\beta$, then the tangent vector v and the corresponding cotangent vector have the same components.

 

[quasar987]

I believe that the words covariant and contravariant refer to the way the components of the vector change with respect to the coordinates system. Suppose $(x^1,\ldots,x^n)$, $(\tilde{x}^1,\ldots,\tilde{x}^n)$ are two intersecting coordinate systems on a manifold M. Suppose for each coordinate system around a point p of M, there is a rule that associates to the coordinates $(x^1,\ldots,x^n)$ of p a set of n numbers (a vector in R^n, then) $(v^1,\ldots,v^n)$.

If the components $v^i$, $\tilde{v}^i$ of the vector corresponding to two coordinate systems $(x^1,\ldots,x^n)$, $(\tilde{x}^1,\ldots,\tilde{x}^n)$ around p are related like so:

$$\tilde{v}^i=\sum_jv_j\frac{\partial x^j}{\partial \tilde{x}^i}$$

then the vector $v=(v^1,\ldots,v^n)$, which we consider the same as the vector $\tilde{v}=(\tilde{v}^1,\ldots,\tilde{v}^n)$, is called a covariant vector.

If, on the other hand, the components are related like so:

$$\tilde{v}^i=\sum_jv_j\frac{\partial \tilde{x}^j}{\partial x^i}$$

then the vector $v=(v^1,\ldots,v^n)$, which we consider the same as the vector $\tilde{v}=(\tilde{v}^1,\ldots,\tilde{v}^n)$, is called a contravariant vector.

So why the names? Probably because the formula as you go from $v$ to $\tilde{v}$ in a covariant vector involves the rate at which $x$ changes with respect to $\tilde{x}$, while in a contravariant vector, it is the contrary: it involves the rate at which $\tilde{x}$ changes with respect to $\tilde{x}$.

Examples:
(1) Suppose we have a curve on an n-manifold M passing through the point p at the time t=0. Then for each coordinate system around p, there corresponds a curve in R^n, and we may differentiate this curve at t=0 to obtain a vector in R^n. If you carry out the computation, you will discover that this is an example of a contravariant vector.

(2) For f a function of a manifold, given a coordinate system around p, you can compute the gradient of the coordinate representation of f at p. This is an example of a covector.

(3) [If you know some classical mechanics] If M=Q is the manifold of physical states of a system and $L:TQ\rightarrow \mathbb{R}$ is a lagrangian function, then for each chart $(q^1,\ldots,q^n)$ of Q (i.e. each set of generalized coordinates) the generalized momenta are defined by

$$p^i:=\frac{\partial L(q^1,\ldots,q^n,v^1,\ldots,v^n)}{\partial v^i}$$

This too defines a covector.


Now, you will often read things like "a contravariant vector is an element of the tangent space and a covariant vector is an element of the cotangent space". What is meant by that is the following. Given a point p on a manifold, we call tangent space at p the vector space $T_pM$ consisting of all linear maps $D:C^{\infty}(M)\rightarrow\mathbb{R}$ satisfying the Leibniz rule "at p" (i.e. D(fg)=D(f)g(p)+f(p)D(g)). It turns out that for a coordinate system $(x^1,\ldots,x^n)$ around p, there is a natural basis for $T_pM$ which we denote (by no accident) $(\partial/\partial x^1|_p,\ldots, \partial/\partial x^n|_p)$. So a general element of $T_pM$ is of the form

$$v=\sum_{i}v^i\left.\frac{\partial}{\partial x^i}\right|_p$$

and the vector $(v^1,\ldots,v^n)$ is contravariant. Indeed, if $(\tilde{x}^1,\ldots,\tilde{x}^n)$ is another coordinate system around p, then by the chain rule

$$v=\sum_{i}v^i\left.\frac{\partial}{\partial x^i}\right|_p=\sum_{i}v^i\left(\sum_j \frac{\partial\tilde{x}^j}{\partial x^i}(p)\left.\frac{\partial}{\partial \tilde{x}^i}\right|_p\right)=\sum_j\left(\sum_iv^i\frac{\partial\tilde{x}^j}{\partial x^i}(p)\right)\frac{\partial}{\partial \tilde{x}^i}\right|_p\right)$$

So, given any contravariant vector $(v^1,\ldots,v^n)$ at p associated to a coordinate system $(x^1,\ldots,x^n)$, you can identify $(v^1,\ldots,v^n)$ with the element

$$\sum_{i}v^i\left.\frac{\partial}{\partial x^i}\right|_p$$

of $T_pM$. Therefor, from the mathematical perspective of structures, the only contravariant vectors at p are the elements of $T_pM$, since any other can be naturally identified with one of these.


Similarly, if you consider $T^*_pM$, the dual space of $T_pM$, and if you note (again, by no accident) $(dx^1_p,\ldots,dx^n_p)$ the basis of $T^*_pM$ dual to the basis $(\partial/\partial x^1|_p,\ldots, \partial/\partial x^n|_p)$ of $T_pM$, then any element of $T^*_pM$ is of the form

$$v=\sum_{i}v_idx^i_p$$

and the vector $(v_1,\ldots,v_n)$ is covariant. Indeed, if $(\tilde{x}^1,\ldots,\tilde{x}^n)$ is another coordinate system around p, then by definition of the differential of a function

$$v=\sum_{i}v_idx^i_p=\sum_{i}v_i\left(\sum_j\frac{\partial x^i}{\partial \tilde{x}^j}d\tilde{x}_p^j\right)=\sum_j\left(\sum_iv_i\frac{\partial x^i}{\partial \tilde{x}^j}\right)d\tilde{x}^j_p$$

So, given any covariant vector $(v_1,\ldots,v_n)$ at p associated to a coordinate system $(x^1,\ldots,x^n)$, you can identify $(v_1,\ldots,v_n)$ with the element

$$\sum_{i}v_idx^i_p$$

of $T^*_pM$. Therefor, from the mathematical perspective of structures, the only covariant vectors at p are the elements of $T^*_pM$, since any other can be naturally identified with one of these.


Tensors of rank (k l) (read "tensor of k contravariant indices and l covariant indices") are defined similarly as a rule associating an array of number $$T^{i_1,\ldots,i_k}_{j_1,\ldots,j_l}$$ to each chart $(x^1,\ldots,x^n)$ around p (where each i and j takes any values between 1 and n) such that if $$\tilde{T}^{i_1,\ldots,i_k}_{j_1,\ldots,j_l}$$ is the array of numbers associated with another coordinate system $(\tilde{x}^1,\ldots,\tilde{x}^n)$, then

$$\tilde{T}^{i_1,\ldots,i_k}_{j_1,\ldots,j_l}=\sum_{i_1'}\ldots\sum_{j_l'}T^{i_1',\ldots,i_k'}_{j_1',\ldots,j_l'}\frac{\partial \tilde{x}^{i_1'}}{\partial x^{i_1}}\ldots\frac{\partial x^{j_l}}{\partial \tilde{x}^{j_l'}}$$

But each of those can be canonically identified with an element of

$$\otimes_{r=1}^kT_pM\otimes \otimes_{s=1}^lT^*_pM$$

so we often say that a tensor of rank (k l) is just an element of the above.