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I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra: $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$, reviewed in good texts (Li & Cheng, etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above.

3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, so you really should be able to see the two dyadics transforming identically, (~), as claimed!

4. This transformation is the form for the transformation of the adjoint, so, then, like ~ (14). I, personally, would not introduce the adjoint this way, and I do not have the book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14), but he does not claim anything of the sort. It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.

5. Not apparent with the information you provide, indeed undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

6. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. So $M^\dagger M$ is invariant under U action. However, imagine a different SU(2) transformation, completely unrelated and oblivious to the above U one, now acting on the right, $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2); and $M^\dagger M$ transforms in the adjoint of the new (global) SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.

I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here, reality of representations, does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra, $[\sigma_j,\sigma_k]=2i\epsilon_{jkl}\sigma_l$, invariant under $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$. This is reviewed in good texts (Li & Cheng, (4.71), etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above.

3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, whence you really should be able to see the two dyadics transforming identically, (~), as claimed!

4. This transformation is the form for the transformation of the adjoint, so, then, like, ~, (14). Personally, I would not introduce the adjoint this way, but I do not have that book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14); but he does not claim anything of the sort. (It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.)

5. Not apparent with the information you provide, indeed, undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

6. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. So $M^\dagger M$ is invariant under U action. However, imagine a different SU(2) transformation, completely unrelated and oblivious to the above U one, now acting on the right, $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2); and $M^\dagger M$ transforms in the adjoint of the new (global) SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.

I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra: $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$, reviewed in good texts (Li & Cheng, etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above.

3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, so you really should be able to see the two dyadics transforming identically, (~), as claimed!

4. This transformation is the form for the transformation of the adjoint, so, then, like ~ (14). I, personally, would not introduce the adjoint this way, and I do not have the book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14), but he does not claim anything of the sort. It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.

5. Not apparent with the information you provide, indeed undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

6. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. However, imagine another SU(2) transformation, completely unrelated and oblivious to the above U one, acting on the right: $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2), and $M^\dagger M$ transforms in the adjoint of the new SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.

I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra: $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$, reviewed in good texts (Li & Cheng, etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above.

3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, so you really should be able to see the two dyadics transforming identically, (~), as claimed!

4. This transformation is the form for the transformation of the adjoint, so, then, like ~ (14). I, personally, would not introduce the adjoint this way, and I do not have the book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14), but he does not claim anything of the sort. It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.

5. Not apparent with the information you provide, indeed undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

6. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. So $M^\dagger M$ is invariant under U action. However, imagine a different SU(2) transformation, completely unrelated and oblivious to the above U one, now acting on the right, $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2); and $M^\dagger M$ transforms in the adjoint of the new (global) SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.

I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

2. It is a long story, and has to do with the conjugacy isomorphisms of the SU(2) Lie algebra, $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$, treated in good books (Li & Cheng, etc..) and several questions on this site. Take it as a given—the "magic" above.

3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, so you really should be able to see the two dyadics transforming identically, (~), as claimed!

4. This transformation is the form for the transformation of the adjoint, so, then, like ~ (14). I, personally, would not introduce the adjoint this way, and I do not have the book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14), but he does not claim anything of the sort. It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.

5. Not apparent with the information you provide, indeed undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

6. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. However, imagine another SU(2) transformation, completely unrelated and oblivious to the above U one, acting on the right: $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2), and $M^\dagger M$ transforms in the adjoint of the new SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.

I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this.

1. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, he is dangerously misleading, in that the "magic" you observe here does not carry through for other SU(N)s. The conjugate representation is equivalent to the fundamental only for SU(2).

2. It is due to the conjugacy isomorphism of the SU(2) Lie algebra: $\zeta^{-1} \sigma^k\zeta =- \sigma^{k~*}$, whence $\zeta^{-1} U\zeta = U^*$, reviewed in good texts (Li & Cheng, etc..) and several questions on this site. Take it as a given if not obvious—the "magic" above.

3. I'm not sure I can second-guess the author's pedagogy. He is demonstrating that $\xi \tilde{\xi}^\dagger$ transforms like (6), $\xi \xi^\dagger$, so you really should be able to see the two dyadics transforming identically, (~), as claimed!

4. This transformation is the form for the transformation of the adjoint, so, then, like ~ (14). I, personally, would not introduce the adjoint this way, and I do not have the book to follow his design. Note this latter matrix H, is not hermitian, $(\xi \tilde{\xi}^\dagger)^\dagger= \tilde{\xi}\xi^\dagger $, unlike the pure dyadic $\xi \xi^\dagger$ and (14), but he does not claim anything of the sort. It is evident that, for (12) to be hermitian, you need the phases of $\xi_1$ and $\xi_2$ to be opposite, so $\xi_1\xi_2$, $\xi_2^2-\xi_1^2$ and $i(\xi_2^2+\xi_1^2)$ to all be real.

5. Not apparent with the information you provide, indeed undecidable. In sympathy with your bafflement, there are superior introductions to SU(2) around, even on this very site, and on MSE. Persistence pays, but if he is not a good match, why, then, do you perseverate?

6. Extra credit. This has presumably nothing to do with your text, but now that one developed the elegant compact language, one would be remiss to not pursue it to the core of the Higgs sector of the SM, the full display of its custodial symmetries. Instead of your dyadic above, bilinear in the spinors $\xi$, one may define another 2×2 matrix, linear in them, whose first column is $\tilde \xi$, and the second $\xi$, $$M\equiv (\tilde\xi , \xi)=\begin{bmatrix}-\xi_2^* & \xi_1 \\ \xi_1^* & \xi_2 \end{bmatrix}~~~\leadsto \\ M^\dagger M = \xi^\dagger \cdot \xi ~~ {\mathbb I} , \hbox{hermitian}. $$ It is evident by above that $$ M'= UM,~~\implies ~~ (M^\dagger M)'= M^\dagger M . $$ since each of its columns transforms thusly. However, imagine another SU(2) transformation, completely unrelated and oblivious to the above U one, acting on the right: $$ M\to MV, $$ for unitary V. The two SU(2)s commute with each other, by inspection, so the group structure is a Cartesian product, SU(2)×SU(2), and $M^\dagger M$ transforms in the adjoint of the new SU(2), V. By suitable squaring and tracings of the hermitian matrix, one may construct the higher-symmetric Higgs potential.