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I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1), on which you had a question; this does not matter, because the U(1) is one of the 6 generators of SO(4), the purpose of my half-answer. It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _\gamma H= i\gamma H,\\ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each component.

The resulting linear action generators in the formalism (2) of 4-vectors is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 -\gamma & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 -\gamma ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -(\gamma +\alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 &\gamma +\alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ is not independent, and could be absorbed in $\alpha_3$. The hypercharge is accounted for in one of the SO(4) generators.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as your $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!

I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1). It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each component.

The resulting linear action generators in the formalism (2) of 4-vectors is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -( \alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 & \alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ could be absorbed in the normalization of $\vec\varphi$.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as your $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!

I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1), on which you had a question; this does not matter, because the U(1) is one of the 6 generators of SO(4), the purpose of my half-answer. It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _\gamma H= i\gamma H,\\ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each components.

The resulting linear action generators in the formalism (2) of 4-vectors, is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 -\gamma & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 -\gamma ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -(\gamma +\alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 &\gamma +\alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ is not independent, and could be absorbed in $\alpha_3$. The hypercharge is accounted for in one of the SO(4) generators.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as you $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!

I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1), on which you had a question; this does not matter, because the U(1) is one of the 6 generators of SO(4), the purpose of my half-answer. It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _\gamma H= i\gamma H,\\ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each component.

The resulting linear action generators in the formalism (2) of 4-vectors is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 -\gamma & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 -\gamma ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -(\gamma +\alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 &\gamma +\alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ is not independent, and could be absorbed in $\alpha_3$. The hypercharge is accounted for in one of the SO(4) generators.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as your $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!

I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1), on which you had a question; this does not matter, because the U(1) is one of the 6 generators of SO(4), the purpose of my half-answer. It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _\gamma H= i\gamma H,\\ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each components.

The resulting linear action generators in the formalism (2) of 4-vectors, is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 -\gamma & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 -\gamma ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -(\gamma +\alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 &\gamma +\alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ is not independent, and could be absorbed in $\alpha_3$. The hypercharge is accounted for in one of the SO(4) generators.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as you $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!

I'll actually not be focused on the explicit symmetry breaking you are asking about, but I'll summarize the full symmetry case (zero mass, or identity mass operator), and summarize the symmetry structure of the symmetric bilinear, $$ \Phi^\dagger \Phi, \tag{1} $$ which covers the case of the kinetic term and diagonal mass. My normalizations will be slightly different/off than in your question, as well as the answers to the questions linked above, including mine, but they do not matter for invariance issues.

The crucial point is that (1) can be rewritten in terms of real 4 vectors $\varphi_i, \quad i = 1,2,3,4$, which display the full SO(4)~SU(2)×SU(2) symmetry structure, but not the "hypercharge" U(1) manifest in (1), on which you had a question; this does not matter, because the U(1) is one of the 6 generators of SO(4), the purpose of my half-answer. It turns out all these symmetries are manifest in the 2×2 non-hermitian matrix H language employed by the cognoscenti, but not in many textbook discussions of the (right) custodial symmetry. (1) can be re-expressed as $$ \vec\varphi\cdot \vec \varphi=\varphi_i \varphi_i, \tag{2} $$ but also as $$ \operatorname{Tr} H^\dagger H = 2 \vec\varphi\cdot \vec \varphi, \tag{3} $$ where $$ H=\varphi_4 +i \varphi_a\sigma ^a= \begin{bmatrix}\varphi_4+i\varphi_3& \varphi_2+i\varphi_1 \\ -\varphi_2+i\varphi_1 & \varphi_4-i\varphi_3 \end{bmatrix}= (\tilde \Phi,\Phi). \tag{4}$$ Yes, I know it looks real weird, but you may identify the doublet in its second column, and the conjugate doublet in its first, while $\phi_1=\varphi_2+i\varphi_1$, and $\phi_2=\varphi_4-i\varphi_3$.

The advantage of the representation (4) is that it is trivial to see that (3) is invariant under the $U(1)\times SU(2)_L\times SU(2)_R$ transformations, $$ H \mapsto e^{i\gamma} e^{i\vec\alpha \cdot \vec\sigma } H e^{i\vec\beta\cdot\vec \sigma}.$$ The over-arrows now represent 3-vectors without risk of confusion, and $\vec \alpha$ the three half-angles of the left weak isospin while the $\vec \beta$ the three half angles of the right-custodial SU(2), visibly commuting with the isospin: the left hand multiplication cares not what the right one is doing. (The half angles incorporate the normalization of the SU(2) generators, so we don't drag 1/2s around.) $\gamma$ is the parameter of the "hypercharge" U(1).

For infinitesimal parameters in the above symmetry map, we see that $$ \delta _\gamma H= i\gamma H,\\ \delta _{\vec \alpha} H= i\vec \alpha\cdot \vec \sigma (\varphi_4+i\vec \varphi\cdot \vec \sigma),\\ \delta _{\vec \beta} H= (\varphi_4+i\vec \varphi\cdot \vec \sigma) i\vec \beta\cdot \vec \sigma, $$which we can evaluate through Pauli matrix identities, and identify the linear action on each components.

The resulting linear action generators in the formalism (2) of 4-vectors, is but the real antisymmetric matrix generating 4-rotations, $$ \begin{pmatrix}0&\beta_3-\alpha_3 -\gamma & \alpha_2-\beta_2&-\alpha_1-\beta_1 \\-(\beta_3-\alpha_3 -\gamma ) &0 &\beta_1-\alpha_1 &-\alpha_2-\beta_2 \\ -\alpha_2+\beta_2 &-\beta_1+\alpha_1 & 0& -(\gamma +\alpha_3+\beta_3) \\\alpha_1+\beta_1 & \alpha_2+\beta_2 &\gamma +\alpha_3+\beta_3 & 0 \\ \end{pmatrix}. \tag{5} $$ Now note these have 6 independent parameters as they should. $\gamma$ is not independent, and could be absorbed in $\alpha_3$. The hypercharge is accounted for in one of the SO(4) generators.

You are now ready to inspect symmetry breaking; however, your (1) and (2) are not equivalent, as you $\varphi_1$ must be degenerate with $\varphi_2$, and $\varphi_3$ with $\varphi_4$: Your hermitian ${\mathbb M}^2$ is 2×2 and only has two eigenvalues!