Basis and dual basis in the Dirac notations In the book An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee, 2nd edition, the author introduced a 'Dirac dictionary':
(left standard notation and right Dirac notation)
$T_i^j e^i \otimes e_j$ relates to  $\sum_{ij} T_{ij} | j \rangle \langle i |$ (I forgot $\sum_{ij}$ in the book previously, editted)
It seems $e^i $  belongs to bra and $e_j$ belones to ket. But, I can upper/lower indices by the metric tensor, e.g., $e^i g_{ij} = e_j$. It looks weird to me if the contraction with the metric tensor will change bra to ket, as $\langle i| g_{ij} = |j \rangle $.
Could the Dirac dictionary be written as, $\sum_{ij} T_i^j e^i \otimes e_j$ relates to $T_{ij} | e_j \rangle \langle e_i |$ and allow $|e^j \rangle$ and $|e^i\rangle$? Thus, $e^i g_{ij} = e_j$ may be related to $\langle e_i| g_{ij} = \langle e^j |$? Looks better to me.
 A: As you noticed, using multiple notations at the same time gets awkward quickly. There is a lot of notational awkwardness here, and the "Dirac dictionary" seems to work only with caveats. (Maybe they are explained in the book, I don't know it.) I will try to clean it up.
Let us start at the beginning. In Einstein notation, we must fix a distinguished basis and the basis vectors are usually called $e_i$. In Dirac notation, we usually don't think of any basis as distinguished, but we may write the vector $e_i$ as $|e_i\rangle$ or just $|i\rangle$. Dirac notation only works if the space is equipped with an inner product, which I will for now write as $(e_i, e_j)$ or $(|i\rangle, |j\rangle)$. In the context where Einstein notation is used, the inner product is usually called the metric and written as $g_{ij} = (e_i, e_j)$.
The next step is to talk about the dual space, which is defined as the space of linear maps from the vector space to $\mathbb C$. It is isomorphic to the vector space, which means that we can identify each dual vector with a "regular" vector. However, there is no canonical isomorphism, i.e., no default choice of this identification. Fixing a basis provides one such identification: in Einstein notation, we define the dual basis vectors $e^i$ to be the linear maps satisfying $e^i(e_j) = \delta^i_j$. Fixing an inner product provides another such identification: in Dirac notation, we define the dual vectors $\langle e_i|$ such that $\langle e_i| \bigl( |e_j\rangle \bigr) = (|e_i\rangle, |e_j\rangle) = g_{ij}$. For convenience, we write that as $\langle e_i | e_j \rangle$ or just $\langle i | j \rangle$. Note that a notation $\langle e^i|$ does not make sense since $\langle v|$ denotes the dual vector of the regular vector $v$, but $e^i$ is not a regular vector.
What we have seen is that $\langle i|$ and $e^i$ are both dual vectors but in general not the same, since $\langle i | j\rangle = g_{ij}$ and $e^i(e_j) = \delta^i_j$. The "dictionary" therefore only works for an orthonormal basis where $g_{ij} = \delta^i_j$ and the position of upper/lower indices does not make a difference.
With this caveat, $T^j_i$ is the same as $T_{ij}$ and $e^i$ the same as $\langle i|$, therefore $T^j_i e^i \otimes e_j = \sum_{ij} T_{ij} |j \rangle \langle i|$. (It is not a good idea to leave out the sums in Dirac notation.) In Dirac notation, we would often just write $\langle j | T | i \rangle$ instead of $T_{ij}$.
Finally, I get to your $e^i g_{ij} = e_j$. Careful: that is not correct! On the left hand side, you have a dual vector; on the right hand side there is a regular vector. They can thus never be equal. The metric tensor can only be used to raise/lower the indices of components to convert between a vector and its dual (using the isomorphism provided by the metric, not the one provided by the dual basis!). To the vector $v = v^i e_i$ we associate the dual vector $v^\flat = v_i e^i$ with $v_i = g_{ij} v^j$. It is a useful exercise to show that $v^\flat(w) = g_{ij} v^j w^i$ using $e^i(e_k) = \delta^i_k$. (If the basis is not orthonormal, the bra $\langle i|$ thus corresponds to $e_i^\flat$ and not to $e^i$!) I hope that solves your remaining questions.
