Can someone help me understand the Euclidean metric? A Euclidean metric is defined as:
$g_{euclid} = g_{ij} = \delta_{ij} dx^i \times dx^j = dx^1dx^1 + \ldots + dx^ndx^n$
Can someone explain the following:


*

*why do we use $dx^i$ instead of $x^i$ which is a coordinate in $R^n$

*Why oes tensor product $\times$ gets turned into multiplication?

*This does not look like a tensor at all, but rather just sum of products...


Finally, a metric is just an inner product. This does not look anything like an inner product! An inner product on Euclidean space is defined as $\langle \cdot , \cdot\rangle$, how is that metric thingy related to my inner product?
 A: *

*Consider the properties of the $dx^i$ compared to the $x^i$. As linear functionals, the basis one-forms $dx^i$ are...well, linear on their arguments. Inner products are bilinear, but we are using two one-forms in each linearly independent term.

*Notational convention. My opinion? It's very misleading. I would almost always keep the tensor product explicit there.

*It's a sum of tensor products, and it defines a symmetric, positive definite bilinear form. Can you see that for any vectors $a,b$, the quantity $g(a,b)$ obeys all the properties of an inner product of $a$ and $b$?

A: I thought I would simply add some details if anyone is wondering how one arrives at the definition of OP.
The Euclidean metric tensor $g$ is a (0,2)-tensor and can therefore be written for two covectors $\omega=\omega_idx^i$ and $\omega'=\omega_i'dx^i$ as
\begin{align}
g=(\omega_idx^i)\otimes(\omega_j'dx^j)=\omega_i\omega_j'dx^i\otimes dx^j,
\end{align}
and since 
\begin{align}
g_{ij}=g(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})=\omega_k\omega_l'(dx^k\otimes dx^l)(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})=\omega_k\omega_l'\langle dx^k,\frac{\partial}{\partial x^i} \rangle \langle dx^l,\frac{\partial}{\partial x^j}\rangle = \omega_i\omega_j'
\end{align}
we see that $g=g_{ij}dx^i\otimes dx^j$. But like you write, $g_{ij}=\delta_{ij}$ and therefore we arrive at
\begin{align} 
g=\sum_{i=1}^n dx^i\otimes dx^i. 
\end{align}
Note that like Muphrid writes, not having the tensor product explicit would be very confusing here.
