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Allen Knutson
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If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.

On thatsuch a component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. Generalize (In the familiar case, $X_C = \{ I/I \}$.) Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^B \cdot \times^B P_{\alpha_k} \times^B X_C \to C$$P_{\alpha_1} \times^I \cdots \times^I P_{\alpha_k} \times^I X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.

(In the familiar case, $X_C = \{ I/I \}$.)

If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.

On that component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^B \cdot \times^B P_{\alpha_k} \times^B X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.

(In the familiar case, $X_C = \{ I/I \}$.)

If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.

On such a component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. (In the familiar case, $X_C = \{ I/I \}$.) Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^I \cdots \times^I P_{\alpha_k} \times^I X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.

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Allen Knutson
  • 27.8k
  • 4
  • 54
  • 152

If $G$ isn't simply connected, then $G(F)$ isn't connected, so $G(F)/I$ isn't connected. Your question is about resolving those $\overline{IwI}/I$ that live in components other than the one containing the basepoint $I/I$.

On that component $C$, there will still be a closed $I$-orbit $X_C$, automatically smooth. Generalize the Bott-Samelson-Demazure-Hansen resolution to $P_{\alpha_1} \times^B \cdot \times^B P_{\alpha_k} \times^B X_C \to C$. The LHS is smooth, and can be made to resolve any choice of $\overline{IwI}/I$, through suitable choice of $\vec \alpha$.

(In the familiar case, $X_C = \{ I/I \}$.)