Mathematicians and physicists use very different languages when they talk about tensors. Fortunately, they are talking about the same thing, but unfortunately, this is not obvious at all. Let me explain.

For simplicity, I'm going to focus on **covariant 2-tensors**, since this case already contains the main intuition. Also, I'm not going to talk about the distinction between covariant and contravariant, but I'll get all the indices right for future study.

## Physicist's definition

Definition: A covariant 2-tensor is a set of numbers $t_{ij}$ with two indices that transforms in a particular way under a change of coordinates… Wait, wait, coordinates in *what space*? Physicists usually don't mention it, but they mean coordinates in a given *vector space* $V$.

More precisely, let $\{\vec e_i\}$ be a basis of the vector space $V$. Then, every **vector** $\vec v$ can be expressed in terms of its *coordinates* $v^i$ as follows:

$$\vec v = \sum_i v^i \vec e_i .$$

So, there are two objects: the vector $\vec v$ which I think of as "solid" or "fundamental", and its coordinates $v^i$, which are "ephemeral", since I have to choose a basis $\vec e_i$ before I can talk about them at all.

Furthermore, in a **different basis** $\{\vec e'_i\}$ of our vector space, the coordinates of one and the same vector $\vec v$ are very different numbers.

$$ \vec v = \sum_i v^i \vec e_i = \sum_i v'^i \vec e'_i .$$

but $v^i ≠ v'^i$. So, the vector is the fundamental thing. Its coordinates are useful for calculations, but they are ephemeral and heavily depend on the choice of basis.

Now, when defining a **covariant 2-tensor**, physicists do something very mysterious: they define a fundamental object (= the 2-tensor) not by describing it directly, but only by specifying how its ephemeral coordinates look like and change when switching to a different basis. Namely, a change of basis

$$ \vec e'_i = \sum_j R_i^a \vec e_a $$

will change the coordinates $t_{ij}$ of the tensor via

$$ t'_{ij} = \sum_{ab} R^a_i R^b_j t_{ab} .$$

If that is not completely unintuitive, I don't know what is.

## Mathematician's definition

Mathematicians define tensors differently. Namely, the give a direct, fundamental description of what a 2-tensor is and only then ponder how it looks like in different coordinate systems.

Here is the definition: a **covariant 2-tensor $t$ is a bilinear map** $t : V\times V \to \mathbb{R}$. That's it. (Bilinear = linear in both arguments).

In other words, a covariant 2-tensor $t$ is a thing that eats two vectors $\vec v$, $\vec w$ and returns a number $t(\vec v, \vec w) \in\mathbb{R}$.

Now, what does this thing look like in coordinates? Choosing a basis $\lbrace \vec e_i \rbrace$, bilinearity allows us to write

$$ t(\vec v, \vec w) = t(\sum_i v^i \vec e_i, \sum_j w^j \vec e_j) = \sum_{ij} v^iw^j t(\vec e_i,\vec e_j) .$$

Now, we simply call the numbers $t_{ij} = t(\vec e_i, \vec e_j)$ the **coordinates of the tensor $t$** in the basis $\vec e_i$. You can calculate that these numbers will behave just like the physicists tell us when you change the basis to $\vec e_i'$. So, the physicist's tensor and the mathematician's tensor are one and the same thing.

## Tensor product

Actually, mathematicians do something more advanced, they define a so called **tensor product** of vector spaces. The previous definition as a bilinear map is still correct, but mathematicians like to write this as "$t\in V^*\otimes V^*$" instead of "$t: V\times V \to \mathbb{R}$ and $t$ bilinear".

However, for a first understanding of the physicist's vs the mathematician's definition, it is not necessary to understand the mathematical tensor product.