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There seem to be two formulations of the Bott–Samelson resolution flowing around. For concreteness, let $ G = \mathrm{GL}_{n} ( \mathbb{C} ) $ with the Borel subgroup $ B \subset G $ of upper triangular matrices, and denote by $ P_i \subset G $ the $i$th parabolic. Fix a reduced expression $ w = s_{i_1} \cdots s_{i_k} $.

Denoting by $ \times^B $ the cross product modulo the diagonal $ B $-action, the first formulation says that multiplication defines a map $$ P_{i_1} \times^B \dotsb \times^B P_{i_k} \xrightarrow{\textrm{mult}} \overline{BwB} \subset G $$ which is a resolution of singularities since it restricts to an isomorphism of open subsets $$ (P_{i_1}-B) \times^B \dotsb \times^B (P_{i_k}-B) \xrightarrow{\;\sim\;} BwB . $$

On the other hand, we can consider the flag variety $ G / B $ and define a map $$ G/B \times_{G/P_{i_1}}\dotsb \times_{G/P_{i_k}} G/B \xrightarrow{\mathrm{first\times last}} \overline{ G ( wB , B ) } \subset G/B\times G/B $$ which drops all coordinates except the first and last one. This is also a resolution of singularities since it restricts to an isomorphism of open subsets $$ \bigl\{\, ( g_0 B , \dotsc , g_nB ) \bigm| g_i B \neq g_{i+1} B \text{ for all $i$} \,\bigr\} \xrightarrow{\;\sim\;} G(wB,B). $$

We note that the RHSs of these maps differ by $ B \backslash G / B \cong (G/B \times G/B)/G $, so the LHSs should differ the same way. For now, let us ignore all stack-related stuff and just pretend that these are honest quotient spaces.

The varieties $ G / P_i $ can be thought of as the varieties of partial flags $ 0 = V_0 \subset V_1 \subset \dotsb \subset V_{i-1} \subset V_{i+1} \subset\dotsb\subset V_n = \mathbb{C}^{n} $ where $ \dim (V_j / V_k) = j - k $. Thus an element in the fibre product $ G/B \times_{ G / P_{i} } G/B $ is of the form $ (gpB,gB) $ for some $ p \in P_i $. Therefore, in passing from one formulation to the other, it is natural to try to write a map like \begin{align*} B\backslash P_{i_1} \times^B \dotsb \times^B P_{i_k}/B & \longrightarrow \bigl( G/B \times_{G/P_{i_1}}\dotsb \times_{G/P_{i_k}} G/B \bigr)/G \\ (p_1 , \dotsc , p_k ) & \longmapsto \bigl( p_1 \dotsm p_k B , p_2 \dotsm p_k B , \dotsc, p_k B , B \bigr) . \end{align*} However, how can this map be well-defined? Any $ (p_1 , \dotsc , p_k ) $ is equivalent to $ (p_1 , \dotsc , p_i b^{-1} , b p_{i-1} , \dotsc , p_k ) $, but I don’t see how $ ( p_1 \dotsm p_k B , p_2 \dotsm p_k B , \dotsc, p_k B , B ) $ can be equivalent to $ ( p_1 \dotsm p_k B , p_2 \dotsm p_k B , \dotsc , b p_{i-1} \dotsm p_{k} B, \dotsc, p_k B , B ) $. What am I doing wrong?

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    $\begingroup$ Does "the $i$th parabolic" mean "the $i$th maximal parabolic", hence presumably the block upper triangular matrices of shape $(i, n - i)$? Also, what does $G(w B, B)$ mean? $\endgroup$
    – LSpice
    Commented Dec 11, 2020 at 19:18
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    $\begingroup$ I don’t know the term “shape $(i,n-i)$.” By the $i$th parabolic, I mean the subgroup of matrices which are almost upper triangular, except that they may be non-zero in the $(i+1,i)$th entry. Is that the same as what you are saying? The notation $G(wB,B)$ means the $G$-orbit containing the element $(wB,B) \in G/B \times G/B$. The $G$-action is the natural left, diagonal action on $G/B \times G/B$. $\endgroup$
    – Gaussler
    Commented Dec 11, 2020 at 19:29
  • $\begingroup$ Ah, you are describing a minimal parabolic, not the maximal parabolic that I guessed (where we allow non-zero entries below the diagonal if both coordinates are at most, or if both coordinates are bigger than, $i$). For example, I would say your minimal parabolic has shape $(1^{i - 1}, 2, 1^{n - i - 1})$. $\endgroup$
    – LSpice
    Commented Dec 11, 2020 at 21:15
  • $\begingroup$ As to $G(w B, B)$, sorry about misunderstanding the notation! I somehow totally missed that it was just the orbit under a group action, and read it instead (nonsensically) as a function $G$ of two arguments. Does your big $\times^B$ product mean that we identify $(g_1, \dotsc, g_k)$ with $(g_1, \dotsc, g_i b, b^{-1}g_{i + 1}, \dotsc, g_k)$ for all $i$ and $B$? (Your later discussion makes it seem so, but that's not what I'm used to calling "the diagonal action"—for example, it's not really a single action of $B$!) $\endgroup$
    – LSpice
    Commented Dec 11, 2020 at 21:19
  • $\begingroup$ Anyway, if I understand correctly, I think it's just a matter of acting on the correct side: note that the element you write doesn't lie in $G/B \times_{G/P_{i_1}} \dotsb \times_{G/P_{i_k}} G/B$, and that that set doesn't naturally admit a right $G$-action. Why not $p \mapsto (B, p_1 B, \dotsc, p_1\dotsm p_k B) \in G\backslash(G/B \times_{G/P_{i_1}} \times \dotsb \times_{G/P_{i_k}} G/B)$? $\endgroup$
    – LSpice
    Commented Dec 11, 2020 at 21:29

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I was being stupid. Of course, as @LSpice also remarks, the real map is \begin{align*} B\backslash P_{i_1} \times^B \dotsb \times^B P_{i_k}/B & \longrightarrow \bigl( G/B \times_{G/P_{i_1}}\dotsb \times_{G/P_{i_k}} G/B \bigr)/G \\ (p_1 , \dotsc , p_k ) & \longmapsto \bigl( B , p_1 B , p_1 p_2 B , \dotsc, p_1 \dotsm p_k B \bigr) . \end{align*}

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    $\begingroup$ @LSpice Of course. Corrected now. Thanks. :-) $\endgroup$
    – Gaussler
    Commented Dec 12, 2020 at 17:06

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