Adjoint, transpose and scalar product

Suppose that a scalar product is defined in the tangent space. If Q 0 is a symmetric, definite positive matrix, it defines a scalar product as:
< p 0 | q 0 > 0 p 0 T Q 0 q 0
where p 0 and q 0 are generic (tangent) vectors (at time 0) and p 0 T is a row vector, the trasnpose of the column vector p 0 . Remark that if the state vector components (then the tangent vector components) have physical dimensions, the components of Q 0 must have inverse square dimensions make the sum of squares possible (the scalar product is then dimensionless).
At time 0 and at time t the scalar product may be different. So, a different matrix Q t is used:
< y t | z t > t y t T Q t z t
Here y t and z t are generic tangent vector at time t .
Let M be a linear operator, like the tangent linear model, applied to tangent vectors at time 0 and giving tangent vectors at time t . By definition its adjoint operator is M such that:
< p t | M q 0 > t < M p t | q 0 > 0
By using the Q matrices, this is written as: p t T Q t M q 0 = ( M p t ) T Q 0 q 0
This is the same as: p t T Q t M q 0 = p t T ( M ) T Q 0 q 0
Since this relation is valid for generic vectors p t and q 0 , it really is a relation between matrices: Q t M = ( M ) T Q 0
Now take the transpose: M T Q t = Q 0 M
The expression of the adjoint is obtained: M = Q 0 -1 M T Q t
From this expression one sees that:
  1. the adjoint intrinsicly depends on the scalar products at initial and final time;
  2. the adjoint is related to the transpose;
  3. the adjoint coincides with the transpose only when both Q s coincide with the identity matrix: Q t = Q 0 = I (in this case both scalar products are Euclidean, or L2); remark that this is possible only if all the state variables have the same physical dimensions.
Remark that the tangent linear operator is applied to tangent vectors at time 0 and gives tangent vectors at time t . Tangent vectors are state variations, or differentials, and are approximated by finite state differences, so they are indicated with δ : δ x t = M δ x 0
The transpose is applied to derivatives at time t and gives derivatives at time 0 , in this sense it goes backward in time: J x 0 = M T J x t
where J=J( x t ) is a generic function of the state at time t .
The adjoint is defined on tangent vectors, not on derivatives (such as the tangent linear model), and goes backward in time (such as the transpose). A transformation between tangent vectors and “derivatives”, which at least accounts for physical dimensions, is then provided by the Q s in the above expression relating adjoint and transpose.
When (since?) what is really needed it the transpose (to compute derivatives with respect to initial conditions), instead of a scalar products one may define a duality form. Tangent vectors and derivatives belong to spaces that are dual to each other and, at time 0 :
( J x 0 | δ x 0 ) ( J x 0 ) T δ x 0 = δ J
At time t :
( J x t | δ x t ) ( J x t ) T δ x t = δ J
These expressions are the same, both coincide with the first variation of J . In the duality form, the transpose behaves like the adjoint does in a scalar product: ( J x t | M δ x 0 )( M T J x t | δ x 0 )
That is to say: ( J x t ) T M δ x 0 = ( M T J x t ) T δ x 0
The duality form (which here appears as a simple product of one row by one column) does not depend on the definition of scalar products (the Q s) and it does not depend on time.
Licenza Creative Commons Francesco Uboldi 2014,2015,2016,2017