Gradient in terms of dyadic product of basis vectors. I am trying to understand what a gradient is expressed as a tensor.
http://web.mit.edu/edbert/GR/gr1.pdf (4.2) says that gradient of a vector or co-variant derivative are same thing.
Then they write $$\tilde{\nabla}\vec{A}=\tilde{\nabla}(A^{\nu}\vec{e}_\nu)=\tilde{e}^{\mu}\partial_{\mu}(A^{\nu}\vec{e}_\nu)=(\partial_{\mu}A^{\nu})\tilde{e}^{\mu}\vec{e}_\nu+A^{\nu}\tilde{e}^{\mu}(\partial_{\mu}\vec{e}_\nu)\equiv(\nabla_{\mu}A^{\nu})\tilde{e}^{\mu}\vec{e}_\nu$$
Similarly for one form;$$\tilde{\nabla}\tilde{A}\equiv(\nabla_{\mu}A_{\nu})\tilde{e}^{\mu}\tilde{e}^\nu$$
This seems little intuitive to me but when i saw the same on wikipedia they have reversed the basis vector order.
https://en.wikipedia.org/wiki/Gradient#Gradient_of_a_vector
Which one is correct? Also I was wondering we can decompose any vector into covariant and contravariant basis so the need for distintction between the two cases not required? I mean $$B=B^{i}e_{i}=B_{i}e^{i}$$ I am not sure about this can someone please confirm and explain?
 A: The expansion ${\bf B}=B^i{\bf e}_i$ can refer only to a vector. An expression like $B_i\tilde{\bf e}^i$ can refer only to a covector (i.e., a 1-form). So we never have $B^i{\bf e}_i=B_i\tilde{\bf e}^i$. However, you can "lower indices" with the metric tensor $g_{ij}$. So (in the presence of a metric) there is a 1-1 correspondence between vectors and covectors.
Now let me dig a little deeper into the two formulas you cite. First, we have two closely related concepts: the covariant derivative of a vector field (say ${\bf A}$) in a direction (say ${\bf u}$), written $\nabla_{\bf u}{\bf A}$. Next, there's the covariant derivative of a vector without specifying the direction: this is written $\nabla{\bf A}$ (except in the Wikipedia article you cite, see below). $\nabla_{\bf u}{\bf A}$ is a vector field (i.e., of type (1,0)) while $\nabla{\bf A}$ is a (1,1) tensor field. The connection is simple: when you contract $\nabla{\bf A}$ with ${\bf u}$, you get $\nabla_{\bf u}{\bf A}$.
I think $\nabla_{\bf u}{\bf A}$ is the more intuitive concept: it says how the vector field ${\bf A}$ is changing in the ${\bf u}$ direction. $\nabla{\bf A}$ is a kind of delayed evaluation: it's a kind of "machine" which has to be fed a direction ${\bf u}$ before coughing up an answer.
Now this is similar to the concepts of directional derivative $\partial_{\bf u}f$ and gradient vector $\nabla f$ from 2nd year calculus:
$$\partial_{\bf u}f = {\bf u}\cdot\nabla f$$
Note the dot product! That's implicit use of the metric. Using components, we'd have to write $\partial_{\bf u}f = g_{ij}u^i\,(\nabla f)^j$. Once people learn about 1-forms, it's more natural to write $\partial_{\bf u}f=u^i\partial_i f$. Here, $\partial_i f$ are the components of a 1-form (in fact, the components of the differential $df$ of $f$). Confusingly, people also call this 1-form the gradient of $f$. The metric coefficients allow us to "pull up" the components: $(\nabla f)^j=g^{ij}\partial_i f$, and $\partial_i f=g_{ij}(\nabla f)^j$.
OK, now about the first formula, $$\nabla{\bf A}=(\nabla_\mu A^\nu)\tilde{\bf e}^\mu{\bf e}_\nu$$
This is saying that if you feed $\nabla{\bf A}$ a direction (say ${\bf u}$), you'll get a vector out, the "directional derivative" of the ${\bf A}$ field in direction ${\bf u}$. To compute the $\nu$-component of that vector, you have to extract the $\mu$-components of ${\bf u}$ (the covector $\tilde{\bf e}^\mu$ does this), multiply it by the number $\nabla_\mu A^\nu$, and sum over $\mu$:
$$\nabla_{\bf u}{\bf A}=(\nabla_\mu A^\nu)u^\mu{\bf e}_\nu$$
Now let's look at the Wikipedia formula:
$$
\nabla {\bf f} =g^{jk}\left(\frac{\partial f^i}{\partial x_j}+\Gamma^i_{\;jl}f^l\right){\bf e}_i{\bf e}_k
$$
Notice the metric $g^{jk}$. Wikipedia is defining a (2,0) tensor by "pulling up" an index from the covariant derivative, which is a (1,1) tensor. Note that Wikipedia calls this a gradient and not a covariant derivative.
We do have a notation conflict between your two sources, with $\nabla {\bf f}$ being used for a (1,1) tensor in the MIT notes, and a (2,0) tensor in that particular Wikipedia article. Also, as I noted, the term "gradient" is sometimes used synonymously with "covariant derivative". Caveat lector!
Almost the simplest example one can give: let $(x,y)$ be the standard coordinate system and let $(x',y')=(2x,2y)$. So the primed grid is twice as fine. The primed basis vectors ${\bf e}'_x,{\bf e}'_y$ are half as long: ${\bf e}'_x=\frac{1}{2}{\bf e}_x$, ditto for $y$. The primed basis covectors however are twice the unprimed: ${\bf \tilde{e}}'_x=2{\bf \tilde{e}}_x$, etc. In terms of Leibniz notation:
$$ {\bf e}'_x = \frac{\partial}{\partial x'} = \frac{\partial}{\partial 2x}
=\tfrac{1}{2} \frac{\partial}{\partial x} = \tfrac{1}{2} {\bf e}_x$$
and
$$ {\bf\tilde{e}}'_x = dx' = d2x = 2\,dx = 2{\bf \tilde{e}}_x$$
The metric coefficients:
$$ g'_{ij}=g({\bf e}'_i,{\bf e}'_j)=
\left[\begin{array}{cc}\tfrac{1}{4}&0\\0&\tfrac{1}{4}\end{array}\right];\quad
{g'}^{ij}=\left[\begin{array}{cc}4&0\\0&4\end{array}\right]
$$
The ${g'}^{ij}$ matrix is the inverse of the $g'_{ij}$ matrix (not the transpose).
Next, let's look at the gradient of $f(x,y)=x+y$. The covariant gradient is just the differential $df=dx+dy=\tfrac{1}{2}(dx'+dy')$. As an invariant geometric object (pictured as a diagonal ruling with parallel lines) $df$ doesn't depend on the coordinate system, but its components in the primed system are half those in the unprimed. The corresponding contravariant gradient is the vector $\text{grad }f$ whose unprimed components are $(1,1)$, and whose primed components are $(2,2)$:
we have the directional derivative $\partial_{\bf u}f=(\text{grad }f)\cdot{\bf u} = df({\bf u})$. The Leibniz formula
$$df = \frac{\partial f}{\partial x'}dx'+\frac{\partial f}{\partial y'}dy'
$$
shows how the halved covariant components go with the doubled basis covectors to give the same $df$. The contravariant gradient $\text{grad }f$ remains the same geometric object (a vector from the origin to the (unprimed) point $(1,1)$ or the (primed) point $(2,2)$), but its primed components are doubled to account for the halved basis vectors.
The story for a vector field ${\bf v}$ is not fundamentally any different. The most "natural" thing to look at is the covariant derivative $\nabla_{\bf u}{\bf v}$, which tells you how fast ${\bf v}$ changes in the ${\bf u}$ direction. This is linear in ${\bf u}$: for any fixed ${\bf v}$, the map ${\bf u}\mapsto\nabla_{\bf u}{\bf v}$ is linear. So (for any fixed  ${\bf v}$) we have a covector, let's denote it $\nabla{\bf v}$, but if we want, we can find a vector, call it $\text{grad }{\bf v}$, such that
$$
\nabla_{\bf u}{\bf v} = (\text{grad }{\bf v})\cdot {\bf u}
$$
This is what the Wikipedia entry is calling the gradient. If the components of $\nabla{\bf v}$ are written $v^i_{;j}$ then the components of $\text{grad }{\bf v}$ are $v^{ik} = g^{jk}v^i_{;j}$.
I think I understand the rationale behind the Wikipedia article. People usually first encounter the gradiant vector, and only later learn about 1-forms, differential forms, and all that. For example, consider the classic physics formula for the force from a potential: $d{\bf p}/dt = -\nabla\Phi$. If $\nabla\Phi$ is interpreted as $d\Phi$, then we have a covector on the right and so we either have to think of momentum as a covector, or else convert: $dp^i/dt=g^{ij}(\partial\Phi/\partial x_j)$. My own preference is to use $\nabla X$ for the covariant derivative (and not the gradient) of $X$, no matter what sort of tensor $X$ is. Note that I have violated this rule above!
As a side issue: the order of certain subscripts is also a matter of convention. The classic GR textbook Gravitation by Misner, Thorne, and Wheeler uses the convention that "the differentiating index comes last": $\nabla_{{\bf e}_i}{\bf e}_j=\Gamma^k_{\;ji}{\bf e}_k$. However, many other sources (for example, the Wikipedia entry you cite) use the opposite convention. 
Finally, Misner, Thorne, and Wheeler spill more ink on more pages just trying to foster intuition about these topics (covariant derivatives, curvature, etc.) than any other source I've seen. It's worth taking a look. (Some people find their style annoying, and some parts of the book are quite dated, but the chapters on differential geometry and tensor calculus are chock-full of very helpful figures.)
