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 Sci Advisor HW Helper PF Gold P: 4,771 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.