Any 'Metric' That Takes a Ket to a Bra? Due to glaring similarities between 'the vectors and one-forms of Riemannian geometry' and 'the kets and bras of linear algebra', I am curious as to whether there exists an object analogous to the metric of the Riemannian geometry (the object that maps a vector to its corresponding one-form) in linear algebra, an object that would map a ket to its corresponding bra. 
I understand that there is a well-defined method of producing a bra corresponding to a ket, namely, taking the adjoint of the column matrix representing the given ket in a particular basis. So, it appears that there might not be a natural need to invoke something similar to the metric of the Riemannian geometry but if we could formulate something like that then, mathematically speaking, we could formulate some other ways of associating a bra with a given ket - just like we can change the metric from the Minkowskian metric to some different metric and the one-form associated with a given vector changes. I am not sure of the advantage of such a thing but I am just curious if there exists any formalism for doing something like that. 
 A: Vectors and 1-forms are not objects of Riemaniann geometry, but of general differential geometry. The tangent and the cotangent bundle are defined as duals of each other, such that at every point of the manifold, the fiber of the cotangent bundle is the dual space of the fiber of the tangent bundle (i.e. the cotangent space at a point is the dual of the tangent space). The metric of (pseudo-)Riemannian geometry provides a canonical way to move between these spaces, but in the finite-dimensional case of manifolds, a vector space and its dual space are always isomorphic. The metric only provides one canonical choice of such an isomorphism.
"Bras" and "kets" typically live in infinite-dimensional Hilbert spaces where the notion of "dual" becomes a bit more subtle, and in particular one would need to distinguish the algebraic dual (all linear functionals) and the continuous dual (all continuous linear functionals) of an infinite-dimensional topological vector space. Often the bras and kets only live in a certain subspace of the full Hilbert space and its continuous dual, which leads to the notion of a rigged Hilbert space.
In any case, for a Hilbert space $H$, it comes equipped with a "metric", namely its inner product, and by the Riesz representation theorem the map
$$ H\to H^\ast, \lvert \psi\rangle \mapsto (\lvert \phi \rangle \mapsto \langle \psi\vert \phi\rangle)$$
is an isomorphism of $H$ and its continuous dual $H^\ast$.
In the end, the question is asking the wrong way around - for vector spaces with an inner product, this notion of a map between the vector space and its dual is natural. (Pseudo-)Riemannian geometry just applies this purely linear algebraic construction at each point of the (co)tangent bundle to the fibers. Of course there are other, non-canonical choices of such maps between the vector space and its dual which could also be expressed as being the natural map for a different inner product, but there is no evident use for that.
A: A Hilbert space is equipped with an inner product $(\ \cdot\ ,\ \cdot\ ) : H\otimes H \to \mathbb C$. One can use it to define a mapping from $H$ to its topological dual $H^*$ (dual pairing) with the assignment
$$\phi\mapsto (\phi,\ \cdot\ ),\qquad\phi\in H,$$
where $(\phi,\ \cdot\ )$ denotes the continuous linear functional $$\psi\mapsto(\phi,\psi),\qquad\forall\psi\in H.$$
A: First a quick review: in general a metric is simply a tensor that takes in two vectors and returns a number, so the 'metric' in quantum mechanics is just the inner product, written as $\langle \psi | \phi \rangle$. Then, just like in relativity, the dual vector (bra) corresponding to a vector (ket) $| \psi \rangle$ is defined to be the function that maps $|\phi \rangle$ to $\langle \psi | \phi \rangle$, which for convenience we write simply as $\langle \psi |$. In both cases a metric/inner product is required to define this map.
Now, how could you change the inner product? One possibility is that you could build your theory off a different product from the start. Orthonormalizing an arbitrary basis will generally yield orthogonal basis vectors with lengths $+1$, $0$, and $-1$. The latter two are problematic because they correspond to zero and negative probabilities, so we can't regard them as physical states. (Indeed in QFT we often have to manually remove these states, e.g. here.) To get valid physical states we always end up with all lengths $+1$, which is exactly the same as the default inner product, so we get nothing new.
(The metric in general relativity is more complex than this because it depends on position, and you can't do this orthogonalization at every single point. In the cases you can, you get $\text{diag}(1, -1, -1, -1)$, i.e. the metric of special relativity.)
Another thing you could do is a passive transformation, which is really just a change of basis. For example, you could replace some normalized basis vector $|0 \rangle$ with $|0 \rangle / 2$, so that the inner product looks different when written out in components. This is useful in relativity because different coordinate systems correspond to different observers. But it's useless in quantum mechanics because there, the inner product gives probabilities, which everybody already agrees on. If I think $|0 \rangle$ is normalized, so will everybody else.
Finally, we could consider an active symmetry transformation. We have to preserve probabilities, and Wigner's theorem states that the most general possibility is a unitary or antiunitary operator. However, this is typically thought of as a transformation of the Hilbert space. While I guess you could think of it as a transformation of the inner product I don't think that gets you anything new.
tl;dr: changing the 'metric' in quantum mechanics isn't useful because all observers agree on probabilities. 
A: Digression on 2-spinors
So this is something that we actually do a lot in 2-spinor notation where we turn a 4-vector into a 2x2 complex Hermitian matrix by adjoining $\sigma_0 = I$ to the usual Pauli matrices $\sigma_{1,2,3}$ and forming $$V = \sum_i v^i~\sigma_i$$ in some basis. The reasons for doing this are rather complicated but let me summarize. One sees rapidly that $\det V = v_\mu~v^\mu$ means that the proper Lorentz transforms take the form $V\mapsto L V L^\dagger$ where $\det L = 1,$ so that this representation realizes directly the connection between special Lorentz transforms and $\text{SL}(2, \mathbb C).$ Null vectors in particular have a special status as projections; $\det V = 0$ implies $V = \phi~\phi^\dagger$, with $\phi \mapsto L\phi$ under a Lorentz transform; the ratio of the components of $\phi$ actually gives a stereographic projection of the direction of $\phi$ on the night sky, down to the complex plane: $L$ is a bilinear transform and we see that Lorentz transforms are all Möbius transforms of the sky. One finally reaps the added bonus that the rotation subgroup $\text{SO}(3)$ is performed by the Lorentz matrix $L = \exp\big((i\theta/2)~\hat n\cdot\vec \sigma\big):$ you also directly see the connection of rotations to the unitary matrices $\text{SU}(2).$ One also sees that the rotation matrix by $2\pi$ is in fact $L=-I$, which does indeed preserve $V$... but it hints that the natural vector space $\mathbb C^2$ that $\phi$ lives in is inherently spinorial: after a full rotation of space everything is mapped to its negative.
There is something like a Lorentz-invariant metric in the 2-spinor space and it is the Levi-Civita matrix $\epsilon_{AB},$ it is not truly a "metric" as it is antisymmetric rather than symmetric, but it can be used canonically to raise and lower indices, and to build the actual metric. But in order to do this, we need a notational way to represent the conjugate space, this idea that we also have $\phi^\dagger \mapsto \phi^\dagger L^\dagger.$
Abstract indices and conjugate spaces
To do this we need the idea of an abstract index: basically a 4-vector (or whatever) really lives in some space $\mathcal V^\bullet$, we make copies of this space for a bunch of symbols and express membership in a space by a superscript; so $v^a$ lives in the copy $\mathcal V^a.$ We identify these vectors by a relabeling isomorphism which we could write as $\delta^a_b$ if you prefer; $v^b$ is the canonically equivalent vector to $v^a$ but living in $\mathcal V^b$ instead of $\mathcal V^a.$ We also identify the covectors $\mathcal V_\bullet$ which are linear maps from $\mathcal V^\bullet$ so some set of scalars; we create copies of this space too and associate it with the corresponding space: $\mathcal V_a$ maps from $\mathcal V^a$ to the scalars. In this way one can encode very precisely and geometrically the underlying relationship with no need for "Einstein summation" etc.
The trick we play here is to produce a copy of $\mathcal V^\bullet$ as a conjugate space, $\bar{\mathcal V}^{\bar\bullet}.$ We specially mark its indices as belonging to the conjugate space and denote the isomorphism as $\big(v^A\big)^\dagger = v^{\bar A};$ we again use the same notation but with a barred index. However our convention is that $$\big(\alpha~u^A + \beta~v^A\big)^\dagger = \alpha^*~u^\bar A + \beta^* v^\bar A.$$The reuse of the same symbol with a bar or without gives some structured meaning by which we can say something like $V$ is Hermitian; we can say that it must have two indices $v^{A\bar A} = \big(v^{A\bar A}\big)^\dagger.$ So when we introduce a spin-basis -- a set $\omicron^A, \iota^A$ such that $\epsilon_{AB} \omicron^A \iota^B = 1$ -- we find that this induces four natural basis vectors for the Hermitian space, $$w^{A\bar A} = \sqrt{\frac12}\left(\omicron^A~\omicron^\bar A + \iota^A~\iota^\bar A\right),\\
x^{A\bar A} = \sqrt{\frac12}\left(\omicron^A~\iota^\bar A + \iota^A~\omicron^\bar A\right),\\
y^{A\bar A} = i\sqrt{\frac12}\left(\omicron^A~\iota^\bar A - \iota^A~\omicron^\bar A\right),\\
z^{A\bar A} = \sqrt{\frac12}\left(\omicron^A~\omicron^\bar A - \iota^A~\iota^\bar A\right).
$$By defining $\alpha$ as a shorthand for the index pair $A\bar A$ we recover our familiar 4-vector indices. We also find the rather astonishing expression for the Minkowski norm, $$\eta_{\alpha\beta} = \epsilon_{AB}~\epsilon_{\bar A\bar B}.$$
How this would apply to quantum mechanics
If we wanted to instead identify a wavefunction $\psi$ as a vector in a Hilbert space $H^\bullet$ we would have to identify a conjugate space $\bar H^\bar \bullet.$ This allows us to canonically identify that $\big(\psi^A\big)^\dagger = \psi^\bar A$ in a consistent way.
The inner product between two wavefunctions would then have the shape $g_{A\bar A}$ as it would connect one wavefunction in conjugate space with one in normal space, $\langle \phi|\psi\rangle = g_{A\bar A}~\psi^A~\phi^\bar A.$ Rather than being symmetric, in this form we have that $g$ is Hermitian and in some sense it lowers indices into a conjugate space rather than their normal one, $g_{A \bar A}~\psi^A = \psi_\bar A.$
I am not sure that this is the right "trick" to play with quantum mechanics but it does lead to density matrices that look like $\psi^A~\psi^\bar A$ in the pure-state case which doesn't look too bad, and it was a very successful trick for the 2-spinor calculus.
