Trying to understand a visualization of contravariant and covariant bases

In general covariant and contravariant vectors live in different spaces $V$ and $V^*$, the tangent space and its dual; the latter being the set of linear maps $f:Vto {mathbb R}$. As a consequence plotting these distinct objects on a single diagram is misleading.

That being said, once we are given an inner product $g({bf x},{bf y})={bf x}cdot {bf y}$ we can use it to identify $V$ with $V^*$ by mapping $fin V^*$ to ${bf f}in V$ through $f({bf x}) = g({bf f},{bf x})$. Given a set of basis vectors ${bf e}_i$ and their inner product ${bf e}_icdot {bf e}_j= g_{ij}$ we can identify two sets of components of a vector ${bf x}in V$ by writing $$ {bf x}= x^1 {bf e}_1+ x^2 {bf e_2}+cdots = x^i {bf e}_i $$ or $$ x_i = {bf e}_icdot {bf x}= g_{ij}x^j $$ In some applications (solid state physics for example) it is useful to introduce vectors ${bf e}^{*i} in V$ such that ${bf e}^{*i} cdot {bf e}_j =delta^i_j$ so that $x^{i}= {bf x}cdot {bf e}^{*i}$ and that is probably what the figure means.

In the absence of an inner product there is still a set of basis functions ${bf e}^{*i}$ for $V^*$ such that ${bf e}^{*i}({bf e}_j)= delta^i_j$, but this is an evaluation of a function and not an inner product.

Some preliminary notions first. Let $V$ be a $k$-vector space, and we write $v = v^i e_ i in V$. Here ${e_i}$ with $i=1,dots,dim V$ is a set of vectors, whereas $v^i$ is a set of elements of the underlying field $k$ which makes $v$ a $k$-linear combination of vectors, thus a vector itself in $V$.

We now introduce the dual $V^vee = mathrm{Hom}(V,k)$ vector space, of functionals from elements of $V$ to $k$. We write $b^i$ for this basis, such that $f = f_i b^i in V^vee$ is an element of the dual vector space.

We have by definition $b^i(e_j) = delta^i_j$, since remember it is a functional and so it takes as input an element of the vector space and spits out an element of $k$.

Now, the metric tensor provides a canonical isomorphism. If I have some $f in V^vee$ there exists a unique $v in V$ such that $f(w) = (v,w)$ for all $w in V$, where $(cdot,cdot)$ is the inner product.

So we have that this $v$ associated to $f$ must satisfy,

$$f(w) = f_i b^i(w^j e_j) = f_i w_j b^i(e_j) = f_i w^j delta^i_j = f_i w^i = g_{ji} v^j w^i$$

and so we see that the required vector $v$ is such that $f_i = g_{ji} v^j$, i.e. they are related by raising or lowering indices. The metric provides a canonical isomorphism.

Now if we have a manifold $M$, with a chart (coordinate system) $phi = (x^1,dots,x^n): U to mathbb R^n$, then for a point $p in U subset M$, we have a basis for the tangent space at $p$, defined by

$$frac{partial}{partial x^i}(f) = (partial_i (f circ phi^{-1}))(phi(p)) $$

for $f in C^infty(M)$. (The formula is a little messy: $f$ is composed with the inverse $phi^{-1}$, then we differentiate it and evaluate it at the point $p$ that is mapped to an element of $mathbb R^n$.)

It would be more intuitive if you look at the illustrations form Daniel A. Fleisch's book on A Student's Guide to Vectors and Tensors.

Here they are:

The parallel projections represent contravariant components of a given vector $vec{A}$ and the perpendicular projections represent the covariant components of the given vector $vec{A}$.

I find the convention used here not quite right. I would opt for the opposite. Since the coordinates can be treated as scalar fields $X^a$, their differentials can be written as $dX^a = e^a_{; i} dx^i = vec{e}^a dvec{x}$ (w.r.t. abstract coordinates $x^i$). Analogously, the gradients w.r.t. $X^a$ can be written as $partial_a = e_{a}^{; i} partial_i = vec{e}_a vecnabla$. Hence, the co-frame $vec{e}^a$ is parallel to the coordinate lines whereas the frame $vec{e}_a$ is orthogonal.

Considering that you're trying to understand a definition, here's a simple example in which you can understand more the definition.

Let's consider 2 vectors, $mathbf{e}_1$ and $mathbf{e}_2$ as they said in the answers (@y255yan) with arbitrary magnitude and directions, and let us consider a vector $mathbf{A}=mathbf{OM}$ represented in this figure

The parallel to the line carrying $mathbf{e}_2$ and passing through $M$ defines another point $M'$, s.t $mathbf{OM'}=x^1mathbf{e}_1$, the same thing with $mathbf{OM''}=x^2mathbf{e}_2$.

Now using the definition: $$mathbf{x}=x^imathbf{e}_i text{Einstein notation}$$

We have the following: $$mathbf{A}=x^1mathbf{e}_1+x^2mathbf{e}_2$$ And here $x^1$ and $x^2$ are the contravariant components od the vector $A$, and to express the covariant components we use the classic definition of scalar product: $$x_1=mathbf{A} . mathbf{e}_1=vertvertmathbf{A}vertvert vertvertmathbf{e}_1vertvert cos alpha x_2= mathbf{A} . mathbf{e}_2=vertvertmathbf{A}vertvert vertvertmathbf{e}_2vertvert sin beta $$

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If the vectors $vertvertmathbf{e}_1vertvert=vertvertmathbf{e}_2vertvert=1$ then the orthogonal projections $m'$ and $m''$ of $M$ represent the covariant components of $mathbf{A}$.