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:

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:

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 ${M}^{\u2020}$ such that:

By using the $Q$ matrices, this is written as: $${p}_{t}^{T}{Q}_{t}M{q}_{0}={\left({M}^{\u2020}{p}_{t}\right)}^{T}{Q}_{0}{q}_{0}$$

This is the same as:$${p}_{t}^{T}{Q}_{t}M{q}_{0}={p}_{t}^{T}{\left({M}^{\u2020}\right)}^{T}{Q}_{0}{q}_{0}$$

Since this relation is valid for generic vectors ${p}_{t}$ and ${q}_{0}$, it really is a relation between matrices:$${Q}_{t}M={\left({M}^{\u2020}\right)}^{T}{Q}_{0}$$

- the adjoint intrinsicly depends on the scalar products at initial and final time;
- the adjoint is related to the transpose;
- 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 $\delta $: $$\delta {x}_{t}=M\delta {x}_{0}$$

The transpose is applied to derivatives at time $t$ and gives derivatives at time $0$, in this sense it goes backward in time: $$\frac{\partial J}{\partial {x}_{0}}={M}^{T}\frac{\partial J}{\partial {x}_{t}}$$

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 $0$:

$$\left(\frac{\partial J}{\partial {x}_{0}}|\delta {x}_{0}\right)\equiv {\left(\frac{\partial J}{\partial {x}_{0}}\right)}^{T}\delta {x}_{0}=\delta J$$

$$\left(\frac{\partial J}{\partial {x}_{t}}|\delta {x}_{t}\right)\equiv {\left(\frac{\partial J}{\partial {x}_{t}}\right)}^{T}\delta {x}_{t}=\delta J$$

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:$$\left(\frac{\partial J}{\partial {x}_{t}}|M\delta {x}_{0}\right)\equiv \left({M}^{T}\frac{\partial J}{\partial {x}_{t}}|\delta {x}_{0}\right)$$

That is to say:$${\left(\frac{\partial J}{\partial {x}_{t}}\right)}^{T}M\delta {x}_{0}={\left({M}^{T}\frac{\partial J}{\partial {x}_{t}}\right)}^{T}\delta {x}_{0}$$