Can the adjoint of the exterior derivative in semi-Riemannian geometry be defined without the Hodge * operator? The adjoint of the exterior derivarive  is defined by
$\delta:=(-1)^k\ast^{-1}d\ast$,
but I need a way which avoids the Hodge $\ast$ operator.

Is there another definition?

For example, for positive definite metric there is the alternative of defining it by
$\int_M \langle d\alpha,\beta\rangle_{k+1} d_{vol} = \int_M\langle\alpha,\delta\beta\rangle_k d_{vol}$
but I don't know if we can apply the Riesz representation theorem to differential forms on semi-Riemannian manifolds, since they do not form a Hilbert space, because the metric is not positive definite.
Can you provide a link or a reference?
 A: An alternative definition of the codifferential is given as formal adjoint of $d$ when this one is expressed using the covariant derivative:
$\delta\omega=-\frac 1 {(k-1)!}\nabla^i\omega_{i i_2 \ldots i_k}d x^{i_2}\wedge\ldots\wedge d x^{i_k}$
I found this formula in Pit-Mann Wong - Einstein Manifolds.
Eventually it turned out that this definition fits my needs.
I wish tot thank to Orbicular, Johannes Ebert and Willie Wong for their help.
I would like to keep the question open for a while, because I am interested in other alternative answers.
A: On any Riemannian manifold, any differential operator has a formal adjoint. 
This has nothing to do with functional analysis, it is a pure calculus fact that uses little more than Stokes theorem. 
This is because taking adjoints is linear, preserves (or rather reverses) 
composition and commutes with multiplication by real-valued functions. 
It is also clear how the adjoint of an order 0 operator (i.e. vector bundle homomorphisms) has to be formed. 
The last and most important step is to construct the adjoint of a vector field on functions. Once this is done,
the existence of an adjoint is proven, because the above operators generate the algebra of differential operator. This last argument also generalizes 
to 
differential operators on vector bundles. The formula below shows you also - for free - that the adjoint is a differential operator, which is 
something you have to work on if you define adjoints on the (pre) Hilbert-space level.
Let $X$ be a vector field, viewed as a differential operator on compactly supported function. To compute the adjoint of $X$, start with the formula
$\int_M Lie_X \alpha = \int_M d \iota_X \alpha + \int_M \iota_x d \alpha =0$ for any $n$-form $\alpha$ and any vector field $X$. The first equality is the Cartan infinitesimal homotopy formula. The second summand is 
zero for degree reasons, the first by Stokes. Now let $\omega$ the volume form and $f,g$ two functions. Therefore
$0= \int_M Lie (fg\omega)= \int_M (Xf) g \omega + \int_M f f (Xg) \omega+ \int_M fg \cdot div (X) \omega$ by the definition of the divergence.
Hence the adjoint of $X$ is $-X - div (X)$. 
So: Once you know how define the volume form on a semi-Riemannian manifold, you know that formal adjoints exists. So you can say that
taking adjoints is a canonical operation on the algebra of differential operators, depending only on the volume form.
The argument above can be easily made into a formula for the adjoint once your operator is given in local coordinates. Whether this formula is useful or enlightening is a matter of taste.
For the exterior derivative, that formula can be written in terms of the Hodge star. This is a (rather simple) theorem. I do not know whether there is any other formula for $d^*$.
Pseudodifferntial operators also have adjoint, which is proven in a different fashion.
It is more difficult to show that the formal adjoint agrees with the Hilbert space adjoint (and I am not really qualified to say much on it and refer you to a book or a real expert)
A: This is no answer, just a too long comment expressing my confusion. I am concerned your claim that one can define the "adjoint" of $d$ in terms of the Riesz representation theorem on Riemannian manifolds. I don't understand that.
Take your favorite compact Riemannian manifold $(M,g)$. Then the exterior derivative is an operator $d:\Omega(M)\rightarrow \Omega(M),$ taking smooth forms to smooth forms. Thus, after suitable Sobolev completions, one can expect $d$ to be a bounded linear map between two Hilbert spaces. 
On the other hand, the Riesz representation theorem concerns bounded linear functionals on a  Hilbert space $H$, i.e. bounded linear maps $H\rightarrow \mathbb{R}.$ It says that any such map is of the form $x\mapsto (x,y)$ for some $y\in H$ (where $(.,.)$ denotes the inner product on $H$). Consequently it is not applicaple to your problem.
I could rather image you are concerned with the theorem on the Hilbert space adjoint. Here one is considering a bounded linear map $A:H_1\rightarrow H_2$ between two different Hilbert spaces. It says there is a bounded linear map $A^*:H_2\rightarrow H_1$ characterized uniquely by the relation $(Ax,y)_2=(x,A^*y)_1$ 
for any $x\in H_1$ and $y\in H_2$.
In your example you want to apply it to the Hilbert space $H_1=H_2=L^2(\Lambda M),$ the space of square-integrable differential forms. Unfortunately, $d$ is not defined on all of $L^2(\Lambda M)$, but only on the space $W^{1,2}(\Lambda M),$ the space of square-integrable differentiable forms whose first derivative is also in $L^2$. Hence $d$ descends to an operator $d:W^{1,2}(\Lambda M)\rightarrow L^2(\Lambda M)$ and its adjoint is an operator $d^*:L^2(\Lambda M)\rightarrow W^{1,2}(\Lambda M)$. But this adjoint is NOT the operator you are looking for!
