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We know that the flag variety $SL(2,\mathbb{C})/B$ which $B$ is Borel subgroup, can be identified with $\mathbb{P^1}$, What can we say about $SL(n,\mathbb{C})/B$ which $B$ is Borel subgroup of $SL(n,\mathbb{C})$

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    $\begingroup$ This is a standard textbook question coming out of the 1950s work of Borel and Chevalley. See the textbooks on linear algebraic groups, or the more recent book by Fulton-Harris (etc.). Beyond the linear algebra description in this special case, the interesting point is that the flag variety is projective (in the general setting of semisimple or reductive groups over any algebraically closed field). In the complex setting it's a compact manifold. $\endgroup$ Oct 8, 2013 at 16:28
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    $\begingroup$ Fulton and Harris discuss this variety and some generalizations in their chapter on homogeneous spaces in their book on Representation Theory. $\endgroup$
    – Ben McKay
    Oct 8, 2013 at 19:20
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    $\begingroup$ When I asked this question, I think I was master student, so if this question was not a research question, I apologize $\endgroup$
    – user21574
    Dec 11, 2017 at 17:23

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The variety $\mathrm{SL}(n,\mathbb{C})/B$ is the variety of complete flags in $\mathbb{C}^n$. That is, a point in the variety can be identified with a chain

$$\{0\}=V_0\subset V_1\subset\dotsc\subset V_{n-1}\subset V_n=\mathbb{C}^n$$

such that $\dim{V_i}=i$. This reduces to $\mathbb{P}^1$ in the case $n=2$, as the only non-proscribed datum at each point is the line $V_1\subset\mathbb{C}^2$.

To prove this, generalizing the $n=2$ result, one checks that the "standard" Borel subgroup stabilizes the "standard" flag

$$\{0\}\subset\langle e_1\rangle\subset\langle e_1,e_2\rangle\subset\cdots\subset\mathbb{C}^n$$

where the $e_i$ are the standard basis vectors.

To expand a bit more, and give an answer that judging from the comments you might like better, let $P$ be a maximal parabolic subgroup of $\mathrm{SL}(n,\mathbb{C})$, so $P$ is proper, contains a Borel, and is maximal with respect to this property. Then $\mathrm{SL}(n,\mathbb{C})/P$ is the Grassmannian of $k$-planes in $\mathbb{C}^n$, for some $k$ depending on $P$, which more naturally generalizes $\mathbb{P}^1$ (the Grassmannian of $1$-planes in $\mathbb{C}^2$). What's happening in the $n=2$ case is that $\mathrm{SL}(2,\mathbb{C})$ is so small that the only parabolic subgroup (up to conjugacy) is Borel.

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    $\begingroup$ @HassanJolany I expanded my answer a bit - hopefully it is now clearer that I am addressing the $n>2$ case. $\endgroup$ Oct 8, 2013 at 15:51
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    $\begingroup$ This answer is correct, so if it isn't the nice variety you are looking for, no one can do anything about your disappointment. For $n=3$, this is the variety of pointed lines in the projective plane, which is the same thing as the variety of complete flags in $\mathbb{C}^2$. $\endgroup$
    – Ben McKay
    Oct 8, 2013 at 19:19
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    $\begingroup$ @HassanJolany If you want an explicit description in coordinates, see the following MathOverflow question: mathoverflow.net/questions/23426/… $\endgroup$
    – S. Carnahan
    Oct 9, 2013 at 1:32
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    $\begingroup$ I understood it to mean that you can get a projective space, but you have to quotient by the right maximal parabolic instead of by a Borel. (I don't know enough to know whether there's any sense in which this parabolic is $U(n-1)$ though). $\endgroup$ Oct 10, 2013 at 18:13
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    $\begingroup$ Matt Pressland@ In fact If we take the intersection of Borel sub-group and and compact real form of $SL(n,\mathbb{C})$, then we will have $U(n-1)$ in this case. Also we can use of this equality $G(k,n)=\frac{U(n)}{U(n-k)\times U(k)}$ .But your answer is OK for me.Thanks $\endgroup$
    – user21574
    Oct 10, 2013 at 19:06

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