For projection matrices, which vector of the outer product should be complex conjugated? I have a conjugation wrong somewhere in my definitions, and I can't work out where it is. I want to define the standard matrix for a projection operator. If you can provide correct and standard conventions such that I get the right answer, then I will accept it as the answer.
Suppose $\{\phi_i\}_i$ and $\{\theta_i\}_i$ are orthonormal bases for $L^2([X, \mathbb{C}], \mu)$. I define the inner product in this way
$$
\langle f,g \rangle = \int_X f \overline g \, d\mu.
$$
Let $f \in L^2$ and for simplicity assume $\langle f,f \rangle = 1$. I define the projection operator of $g \in L^2$ on $f$ in this way:
$$
P^fg = \langle g, f\rangle f.
$$
With these definitions, my conjugation becomes wrong when I want to find the standard matrix $A$, for the projection operator $P^f$ relative to the bases $\{\phi_i\}_i$ on the domain, and $\{\theta_i\}_i$ on the co-domain. I try:
$$
A_{ij} = \langle \theta_i , P^f \phi_j\rangle = \langle \theta_i , \langle\phi_j, f\rangle f \rangle = \langle \theta_i ,  f \rangle \overline{ \langle\phi_j, f\rangle } = \overline{\langle f, \theta_i  \rangle } \langle f, \phi_j \rangle.
$$
With this conclusion, I would believe that $A = \overline{c} d^T$ where $c$ is the vector with elements $c_i = \langle f, \theta_i \rangle$ and $d_i = \langle f, \phi_i \rangle$ for every $i$. But I recall seeing this expression in text books, and the expression should have the conjugate on $d$ in stead of $c$. It should read $A = c d^\dagger$. Can you see which convention I got wrong since I have the wrong answer? In case there are multiple ways to do it, provide the conventions that you feel is most common in mathematics (not in physics this time).
 A: As Gert Pedersen writes in Chapter 3 of his renowned book "Analysis NOW":

"A mathematical physicist is a mathematician believing that a sesquilinear form is conjugate linear in the first variable and linear in the second."

What this means is that -- contrary to mathematics -- in mathematical physics the inner product $\langle\cdot,\cdot\rangle$ on any (complex) Hilbert space is defined to be conjugate linear in the first and linear in the second argument. You can verify this by checking out any book on quantum mechanics, e.g., Chapter II in the standard work of Reed & Simon ("Methods of Modern Mathematical Physics I: Functional Analysis") to name just one.
Now the inconsistency in your calculation is that you define your inner product and your projection operator "like a mathematician" but the matrix representation of $P^f$ "like a physicist". More precisely, the operator always has to go where the inner product is linear ${}^\text{footnote 1}$.
As a consequence, switching convention of where the inner product is linear amounts to complex conjugation of the representation matrix ($A\to\overline{A}$). This is where your missing complex conjugate went.
Let us use this insight for determining the representation matrix of $P_f$ for either convention.
First note that the orthogonal projection $P^f$ onto the linear subspace spanned by $f$ is given by $ff^\dagger$ (or as a physicist would write, $|f\rangle\langle f|$) where $f^\dagger$ is the linear functional induced by $f$ via the Riesz representation theorem, i.e. $P^f(x)=(ff^\dagger)(x)=f^\dagger(x)\cdot f$ where $f^\dagger(x)$ is a complex number.
Also this notation is independent of any inner product convention, cf. this discussion of different definitions of inner products in different fields.
Now for the computation.

*

*Representation matrix for "mathematicians": From footnote 1 we know that the representation matrix of $P^f$ is given by $(\langle P^f\phi_j,\theta_i\rangle)_{i,j}$, that is,
$$
\langle P^f\phi_j,\theta_i\rangle=\big\langle \big(\langle\phi_j,f\rangle f\big)\theta_i\big\rangle=\langle\phi_j,f\rangle\langle f,\theta_i\rangle=\langle f,\theta_i\rangle\overline{\langle f,\phi_j\rangle}\,.$$

*Representation matrix for "physicists": again by footnote 1 the entries of the representation matrix of $P^f$ read
$$
\langle \theta_i,P^f\phi_j\rangle=\big\langle\theta_i,\big(\langle f,\phi_j\rangle\big) f\big\rangle=\langle f,\phi_j\rangle\langle\theta_i,f\rangle=\overline{\langle f,\theta_i\rangle}\langle f,\phi_j\rangle\,.
$$
This recovers the result from footnote 1 that a switch in convention amounts to complex conjugation of the representation matrix.

${}^\text{footnote 1}\,$: To verify this let us recall some basic linear algebra: we start with a linear map $T$ between (finite-dimensional, complex) inner product spaces $\mathcal H_1,\mathcal H_2$ and orthonormal bases $\{\phi_j\}_j$ of $\mathcal H_1$ and $\{\theta_i\}_i$ of $\mathcal H_2$.
By the basis property there for each $j$ exist unique coefficients $a_{ij}$ such that
$
T\phi_j=\sum_i a_{ij}\theta_i
$,
and $A:=(a_{ij})_{i,j}$ is called representation matrix of $T$.
Because we have access to an inner product we can even determine these $a_{ij}$:
$$
\langle T\phi_j,\theta_i\rangle=\Big\langle\sum_{i'} a_{i'j}\theta_{i'},\theta_i\Big\rangle=\sum_{i'} \big\langle a_{i'j}\theta_{i'},\theta_i\big\rangle
$$
This is where convention comes into play; if the inner product is linear in the first argument ("mathematician"), then
$$
\langle T\phi_j,\theta_i\rangle=\sum_{i'} \big\langle a_{i'j}\theta_{i'},\theta_i\big\rangle=\sum_{i'} a_{i'j}\underbrace{\langle \theta_{i'},\theta_i\rangle}_{i'=i}=a_{ij}
$$
so the matrix representation of $T$ is given by $(a_{ij})_{i,j}=(\langle T\phi_j,\theta_i\rangle)_{i,j}$. However, if the inner product were linear in the second argument ("physicist"), then
$$
\langle T\phi_j,\theta_i\rangle=\sum_{i'} \big\langle a_{i'j}\theta_{i'},\theta_i\big\rangle=\sum_{i'} \overline{a_{i'j}}\underbrace{\langle \theta_{i'},\theta_i\rangle}_{i'=i}=\overline{a_{ij}}
$$
meaning $a_{ij}=\overline{\langle T\phi_j,\theta_i\rangle}=\langle\theta_i,T\phi_j\rangle$. Either way the "location" of $T$ corresponds to where the inner product is assumed to be linear, and the difference between the two conventions manifests in an additional complex conjugation of the $a_{ij}$.
