Show that $\sum_{i=1}^m\left(D(g_i){\bf{x}}, D(g_i){\bf{y}}\right)=\left(\sum_{i=1}^m D^{\dagger}(g_i)D(g_i){\bf{x}},{\bf{y}}\right)$

Question

I am reading "Applications of group theory in quantum mechanics" by M.I. Petrashen & E.D. Trifonov. In chapter 3, section 3.4 - Existence of an equivalent unitary representation, on page 29, I am stuck on equation (3.32) which states that $$\sum_{i=1}^m\left(D(g_i){\bf{x}}, D(g_i){\bf{y}}\right)=\left(\sum_{i=1}^m D^{\dagger}(g_i)D(g_i){\bf{x}},{\bf{y}}\right)\tag{3.32}$$ Petrashen clearly states that the $D(g_i)$ are representation matrices and denotes the scalar product as $$({\bf{x}},{\bf{y}})=x_1\overline{y_1}+ x_2\overline{y_2}+\cdots + x_n\overline{y_n}\tag{A}$$ Where $\overline{y_i}$ is (presumably) the complex conjugate, ${y_i}^\ast$.

The sum of scalar products is, $$\sum_{i=1}^m\left(D(g_i){\bf{x}}, D(g_i){\bf{y}}\right)$$

$$=D(g_1)x_1 D^\ast(g_1){y_1}^{\ast}+D(g_2)x_2D^\ast(g_2){y_2}^{\ast}+\cdots +D(g_m)x_mD^\ast(g_m){y_m}^{\ast}\tag{B}$$ I don't understand how equation $(3.32)$ was deduced, as taking the Hermitian adjoint/conjugate of both sides of $(\mathrm{B})$ yields,

$$\left(\sum_{i=1}^m\left(D(g_i){\bf{x}}, D(g_i){\bf{y}}\right)\right)^\dagger$$

$$=\Big(D(g_1){x_1}D^\ast(g_1){y_1}^{\ast}\Big)^{\dagger}+\Big(D(g_2)x_2D^\ast(g_2){y_2}^{\ast}\Big)^\dagger+\cdots +\Big(D(g_m)x_mD^\ast(g_m){y_m}^{\ast}\Big)^\dagger$$

$$=(D^\dagger (g_1){x_1}^\ast {D^\ast}^{\dagger}(g_1) y_1+ D^\dagger (g_2){x_2}^\ast {D^\ast}^{\dagger} (g_2)^\dagger y_2+\cdots + D^\dagger (g_m){x_m}^{\ast} {D^\ast}^{\dagger} (g_m) y_m$$ I don't know how to proceed from here to show that this is equal to

$$\left(\sum_{i=1}^m D^{\dagger}(g_i)D(g_i){\bf{x}},{\bf{y}}\right)$$

I am reading another book by Michael Tinkham called "Group Theory and Quantum Mechanics", which in pdf form can be found here and in Appendix A, page 317 (or page 327 if viewing online/electronically), Tinkham denotes the inner product or Hermitian scalar product as $$({\bf{v}},{\bf{w}})=\sum_k v_k^\ast w_k.$$ For reasons I don't understand the complex conjugation is in the opposite order to the way in which Petrashen denotes the scalar product in $(\mathrm{A})$ (where the second vector is complex conjugated). but even with that adjustment, this is still not equal to the RHS of $(3.32)$, $$\left(\sum_{i=1}^m D^{\dagger}(g_i)D(g_i){\bf{x}},{\bf{y}}\right)$$

I'm missing something here, can someone please explain how I can show that $$\sum_{i=1}^m\left(D(g_i){\bf{x}}, D(g_i){\bf{y}}\right)=\left(\sum_{i=1}^m D^{\dagger}(g_i)D(g_i){\bf{x}},{\bf{y}}\right)\tag{3.32}?$$

Answer 1

I didn't look up the reference, but presumably the Hermitian form is defined so that $\langle u, Mv\rangle = \langle M^\dagger u, v\rangle$ for any matrix $M$. In that case you just have $$\sum_{i=1}^m \langle D(g_i) \mathbf{x}, D(g_i) \mathbf{y}\rangle = \sum_{i=1}^m \langle D(g_i)^\dagger D(g_i) \mathbf{x}, \mathbf{y}\rangle = \left\langle \sum_{i=1}^m D(g_i)^\dagger D(g_i) \mathbf{x}, \mathbf{y}\right\rangle.$$

Your equation (B) does not make sense. The use of $x_i$ and $y_i$ is not appropriate.

Whether a Hermitian form is conjugate-linear in the first or second argument is inconsistent across the literature. You'll often have to reverse the order to fix things up to be compatible across sources.

Edit in response to follow-up comment: The identity $\langle u, Mv\rangle = \langle M^\dagger u, v\rangle$ is the usual abstract definition of $M^\dagger$, where $\langle -, -\rangle$ is a fixed non-degenerate Hermitian form. It's very easy to find sources that take this view, like the Wikipedia article.

You can also prove it directly using the usual coordinate-based definition. If $(M^\dagger)_{ij}$ is defined to be $\overline{M_{ji}}$, then:

\begin{align*} \langle u, Mv\rangle &= \sum_{i=1}^n u_i \overline{(Mv)_i} \\ &= \sum_{i=1}^n u_i \overline{\sum_{j=1}^n M_{ij} v_j} \\ &= \sum_{i=1}^n u_i \sum_{j=1}^n \overline{M_{ij}} \overline{v_j} \\ &= \sum_{i=1}^n u_i \sum_{j=1}^n (M^\dagger)_{ji} \overline{v_j} \\ &= \sum_{i=1}^n \sum_{j=1}^n u_i (M^\dagger)_{ji} \overline{v_j} \\ &= \sum_{j=1}^n (\sum_{i=1}^n (M^\dagger)_{ji} u_i) \overline{v_j} \\ &= \sum_{j=1}^n (M^\dagger u)_j \overline{v_j} \\ &= \langle M^\dagger u, v\rangle. \end{align*}