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This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).

One should be careful speaking casually of the Levi component of a linear algebraic group, since, over general fields, not all groups have to have them, and, when they do have them, they need not be rationally conjugate. (You're working over $\mathbb C$, but the same question could be asked over any field $k$, and the answer is the same. I'll work in that generality.) A parabolic subgroup of a reductive group does always have a Levi component, and they're all rationally conjugate, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is usually not a parabolic subgroup of $G$.

Of course, in this case, $P'$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$ [BT, Proposition 4.4], its Levi component $M$ containing $T$ is also a Levi component of $P'$ [1], and all Levi components of $P'$ arise in this way [BT, Proposition 4.7]. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected [2].

[BT]: Borel and Tits - Groupes réductifs.

[1]: This may be verified over the algebraic closure, so assume that $k$ is algebraically closed. Let $U'$ be the unipotent radical of $P'$. Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$. Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'\cdot U_1$.

[2]: This is folklore, but I don't know a good published reference. At their answer to Centralizers of subtori in reductive groups, derived subgroups, @nfdc23 points out Corollary 9.5.11 of Conrad - Reductive groups over fields.

This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).

One should be careful about speaking casually of the Levi component of a linear algebraic group, since, over general fields, not all groups have one, and, when a group has one, the various possible components need not be rationally conjugate. (You're working over $\mathbb C$, where this problem doesn't arise, but the same question could be asked over any field $k$, and the answer is the same. I'll work in that generality.) A parabolic subgroup of a reductive group does always have a Levi component, and they're all rationally conjugate, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is usually not a parabolic subgroup of $G$.

Of course, in this case, $P'$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$ [BT, Proposition 4.4], its Levi component $M$ containing $T$ is also a Levi component of $P'$ [1], and all Levi components of $P'$ arise in this way [BT, Proposition 4.7]. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected [2].

[BT]: Borel and Tits - Groupes réductifs.

[1]: This may be verified over the algebraic closure, so assume that $k$ is algebraically closed. Let $U'$ be the unipotent radical of $P'$. Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$. Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'$.

[2]: This is folklore, but I don't know a good published reference. At their answer to Centralizers of subtori in reductive groups, derived subgroups, @nfdc23 points out Corollary 9.5.11 of Conrad - Reductive groups over fields.

This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).

One should be careful speaking casually of the Levi component of a linear algebraic group, since, over general fields, not all groups have to have them, and, when they do have them, they need not be rationally conjugate. (You're working over $\mathbb C$, but the same question could be asked over any field $k$, and the answer is the same. I'll work in that generality.) A parabolic subgroup of a reductive group does always have a Levi component, and they're all rationally conjugate, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is not a parabolic subgroup.

Of course, in this case, $P'$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$ [BT, Proposition 4.4], its Levi component $M$ containing $T$ is also a Levi component of $P'$ [1], and all Levi components of $P'$ arise in this way [BT, Proposition 4.7]. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected [2].

[BT]: Borel and Tits - Groupes réductifs.

[1]: This may be verified over the algebraic closure, so assume that $k$ is algebraically closed. Let $U'$ be the unipotent radical of $P'$. Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$. Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'\cdot U_1$.

[2]: This is folklore, but I don't know a good published reference. At their answer to Centralizers of subtori in reductive groups, derived subgroups, @nfdc23 points out Corollary 9.5.11 of Conrad - Reductive groups over fields.

This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).

One should be careful speaking casually of the Levi component of a linear algebraic group, since, over general fields, not all groups have to have them, and, when they do have them, they need not be rationally conjugate. (You're working over $\mathbb C$, but the same question could be asked over any field $k$, and the answer is the same. I'll work in that generality.) A parabolic subgroup of a reductive group does always have a Levi component, and they're all rationally conjugate, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is usually not a parabolic subgroup of $G$.

Of course, in this case, $P'$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$ [BT, Proposition 4.4], its Levi component $M$ containing $T$ is also a Levi component of $P'$ [1], and all Levi components of $P'$ arise in this way [BT, Proposition 4.7]. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected [2].

[BT]: Borel and Tits - Groupes réductifs.

[1]: This may be verified over the algebraic closure, so assume that $k$ is algebraically closed. Let $U'$ be the unipotent radical of $P'$. Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$. Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'\cdot U_1$.

[2]: This is folklore, but I don't know a good published reference. At their answer to Centralizers of subtori in reductive groups, derived subgroups, @nfdc23 points out Corollary 9.5.11 of Conrad - Reductive groups over fields.

This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).

One should be careful speaking casually of the Levi component of a subgroup, since linear algebraic groups over general fields don't have to have them. (You're working over $\mathbb C$, but the same question could be asked over any field $k$, and the answer is the same. I'll work in that generality.) A parabolic subgroup of a reductive group does always have a Levi component, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is not a parabolic subgroup.

Of course, in this case, $P'$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$ [BT, Proposition 4.4], and its Levi component $M$ containing $T$ is also a Levi component of $P'$ [1]. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected.

[BT]: Borel and Tits - Groupes réductifs

[1]: This may be verified over the algebraic closure, so assume that $k$ is algebraically closed. Let $U'$ be the unipotent radical of $P'$. Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$. Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'\cdot U_1$.

This is an answer to the modified question, although, as @KentaSuzuki suggested, it might be better just to ask this question separately (in which case I am happy to move my answer to a different question).

One should be careful speaking casually of the Levi component of a linear algebraic group, since, over general fields, not all groups have to have them, and, when they do have them, they need not be rationally conjugate. (You're working over $\mathbb C$, but the same question could be asked over any field $k$, and the answer is the same. I'll work in that generality.) A parabolic subgroup of a reductive group does always have a Levi component, and they're all rationally conjugate, but $P' \mathrel{:=} P_1 \cap w P_2 w^{-1}$ is not a parabolic subgroup.

Of course, in this case, $P'$ does have a Levi component, but it's worth saying why. Probably the easiest way to see why is that $P'\cdot U_1$, where $U_1$ is the unipotent radical of $P_1$, is a parabolic subgroup of $G$ [BT, Proposition 4.4], its Levi component $M$ containing $T$ is also a Levi component of $P'$ [1], and all Levi components of $P'$ arise in this way [BT, Proposition 4.7]. In this context, the derived subgroup of $M$ is simply connected, because the derived subgroup of a Levi component of a parabolic subgroup of a simply connected group is always simply connected [2].

[BT]: Borel and Tits - Groupes réductifs.

[1]: This may be verified over the algebraic closure, so assume that $k$ is algebraically closed. Let $U'$ be the unipotent radical of $P'$. Then $U'\cdot U_1$ is smooth, connected, unipotent, and normal in $P'\cdot U_1$, and $(P'\cdot U_1)/(U'\cdot U_1) \cong P'/U'$ is reductive, so $U'\cdot U_1$ is the unipotent radical of $P'\cdot U_1$. Therefore, the projection from $P'/U' \cong (P'\cdot U_1)/(U'\cdot U_1)$ onto $M$ is an isomorphism, so $M$ is a Levi component of $P'\cdot U_1$.

[2]: This is folklore, but I don't know a good published reference. At their answer to Centralizers of subtori in reductive groups, derived subgroups, @nfdc23 points out Corollary 9.5.11 of Conrad - Reductive groups over fields.