Which is the difference between the delta tensor $\delta_{a}^{b}$ and the metric tensor $g_{ab}$? I don't understand which is the difference between the delta of Kronecker $\delta_{a}^{b}$ and the metric tensor $g_{ab}$. They looks to have the same effect when raising and lowering the indices, but the former should be the identity, while the latter is a dynamical object.
Can anyone help me?
 A: This is where you need to start thinking about tensors as multi-linear maps.
A tensor with two down indices can take two vectors and give a scalar: $g_{\mu\nu} x^\mu y^\nu$ is a scalar. Thus, a tensor $g_{\mu\nu}$ is a multilinear map $V \times V \to \mathbb R$. The tensor product is in fact defined such that every multilinear map $V \times V \to \mathbb R$ is associated with exactly one tensor with two down indices.
Now, exploiting partial application we realize that we can feed a multilinear map $V \times V \to \mathbb R$ a vector and get a linear map $V \to \mathbb R$. In index notation, $g_{\mu\nu} x^\mu = y_\nu$. The space of linear maps $V \to \mathbb R$ is denoted by $V^*$. So, by partial application we can view a tensor $g$ as either of type $V \times V \to \mathbb R$ or as of type $V \to (V \to \mathbb R) = V \to V^*$. 
As a linear map $V \to V^*$, $g$ can be invertible. If it is, which I will be assuming henceforth, then $g^{-1}$ is $V^* \to V$. If $V$ is finite dimensional, then $V = (V^*)^* = (V^* \to \mathbb R)$. We can thus reverse the argument above to view $g^{-1}$ as a multilinear map $V^* \times V^* \to \mathbb R$. In indices, $g^{-1}$ is a tensor with two up indices, $(g^{-1})^{\mu\nu}$. Because $g^{-1}$ is unique if it exists, and the type (indicated by the indices)makes clear which tensor we are talking about, we might as well write $g^{\mu\nu}$. 
Because $g_{\mu\nu}$ and its inverse define linear maps $V \leftrightarrow V^*$, for any multilinear map that takes any number of arguments from $V$ or $V^*$ -- i.e., any tensor -- we can freely swap $V$-arguments and $V^*$ argument. Recall that a $V$-argument is the same as a down index, and a $V^*$ argument is the same as an up index; this means that we can freely raise and lower indices on arbitrary tensors. Because this swapping is unique, owing to the invertibility of $g$, we denote them with the same kernel letters, e.g., $C^\mu{}_{\nu\kappa \lambda}, C^\mu{}_\nu{}^\kappa{}_\lambda$.
The Kroenecker $\delta$ is the tensor corresponding to the identity map $I : V \to V$ or $I : V^* \to V^*$. The difference compared to the metric or its inverse is in it the type. The Kroenecker is $V \times V^* \to \mathbb R$; the metric is $V \times V \to \mathbb R$. Of course, these can be freely swapped as we just discussed, so there is a $\delta_{\mu\nu}$ tensor, but you can realize that $\delta_{\mu\nu} = g_{\mu\nu}$, and likewise for the up index variants. From this, we can also write $g^\mu{}_\nu = \delta^\mu{}_\nu$, if we wish. But when doing so, we should remember that this object is more or less $gg^{-1}$, which explains why it doesn't actually depend on $g$.
In Euclidean flat space in Cartesian coordinates and in Euclidean flat space in Cartesian coordinates only, are the components equal, in the sense that $g_{\mathfrak{a}\mathfrak{b}} = \delta_{\mathfrak a}{}^{\mathfrak b} = \begin{cases}1 & \mathfrak a = \mathfrak b \\ 0 & \text{otherwise} \end{cases} $.1 In any other coordinates or frames, or with Lorentzian signature, or any curved space, $\delta_{\mathfrak a}{}^{\mathfrak b}$ will be the same, but $g_{\mathfrak{a}\mathfrak{b}}$ can be anything. (Subject to constraints like $g$ being symmetric, invertible, of a certain signature.) In those cases, it is imperative to keep track of whether your indices are up or down.
The bottom line is that tensor products and metrics don't make sense unless thought of in the context of multilinear maps.2 Misner, Thorne, and Wheeler emphasize this in their definitions of various physical tensors, so it is not all formal mathematics. In fact, the "physicist" definition of a tensor as a set of numbers that transforms in a certain way is awful and doesn't tell you what a tensor does, physically. A tensor, to use MTW's phrase, is a linear machine that you feed vectors to get scalars. It tells you how some scalar quantity depends on several vectorial quantities. 

1 The fraktur here is because I am using abstract index notation, which is how to implement the view above of tensors computationally. See for instance the Geroch-Held-Penrose paper or Penrose and Rindler, Spinors and Space-Time.
2 And to fully appreciate tensors, you must think about the tensor product as a functor.)
A: The metric tensor $g_{\mu\nu}$ has two lower indices. The Kronecker delta $\delta^\mu_\nu$ has one lower and one upper index.  (It's never found with two upper or two lower indices in covariant expressions, so it can't be used to raise or lower indices.)
A: They are quite different objects. The Kronecker delta produces an identity-like quantity, while the metric tensor is dependent on your space. For regular flat 3D space, you have the metric tensor as $g^{ab} = g_{ab} = \rm{diag}(1,1,1)$, but that will be different for different cases. Minkowski spacetime is represented by $g^{\mu \nu} = g_{\mu \nu} = \rm{diag}(-1,1,1,1)$, the surface of a sphere is represented by $g_{ab} = \rm{diag}(R^2, R^2sin^2\theta)$, but with dual metric $g^{ab} = \rm{diag}(R^{-2}, \left[Rsin\theta\right]^{-2})$.
There are infinitely many metrics and many examples are out there, but only a single Kronecker delta. You should also notice that the metric tensor and its dual can be used to raise and lower indices, while the Kronecker delta is used as a means to calculate the trace of a tensor.
A: We can use the tensor $g_{\mu\nu}$ and its inverse $g^{\mu\nu}$ to lower and raise indices on tensors. In particular a two-index tensor satisfies $$g_{\mu\nu}X^{\nu\rho}=X_\mu^{\:\rho},g_{\mu\nu}X^{\rho\nu}=X^\rho_{\:\mu},\,g_{\rho\sigma}X_\mu^{\:\rho}=X_{\mu\sigma},\,g_{\rho\sigma}X^\rho_{\:\mu}=X_{\sigma\mu}$$and similarly identities using $g^{\mu\nu}$. The special case $X=g$ gives $g_{\mu\nu}g^{\nu\rho}=g_\mu^{\:\rho}$. But when we say one metric tensor is the other's inverse, we mean simply that $g_\mu^{\:\rho}=\delta_\mu^{\:\rho}$. However, in a general manifold the Kronecker delta's entries are only obtained for $g$ if it is mixed (i.e. it has a lower and an upper index). Any invertible symmetric matrix can be chosen for $g_{\mu\nu}$: the parity of its number of negative eigenvalues is determined by the geometry, and most other degrees of freedom are a coordinate choice. (One cannot, however, use a coordinate transformation to change whether the Riemann tensor is identically zero, which somewhat constrains $g_{\mu\nu}$.)
