<p>Let <i>x</i> be any nilpotent endomorphism of a vector space <i>V</i> of finite dimension over an algebraically closed field of arbitrary characteristic. The Springer fiber <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(\mathcal {F}_x\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="script">F</mi> <mi>x</mi> </msub> </math></EquationSource> </InlineEquation> is the subset of <i>x</i>-stable complete flags. In the case <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(x^2=0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, the components of <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(\mathcal F_x\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="script">F</mi> <mi>x</mi> </msub> </math></EquationSource> </InlineEquation> are parameterized by Young tableaux of shape <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\((2^k,1^{n-2k})\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <msup> <mn>2</mn> <mi>k</mi> </msup> <mo>,</mo> <msup> <mn>1</mn> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mi>k</mi> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> of two columns, where <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(k=\textrm{Rank}\,x\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>k</mi> <mo>=</mo> <mtext>Rank</mtext> <mspace width="0.166667em" /> <mi>x</mi> </mrow> </math></EquationSource> </InlineEquation>. In this paper, we present an equivalent parameterization of the components of <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\(\mathcal F_x\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="script">F</mi> <mi>x</mi> </msub> </math></EquationSource> </InlineEquation>, and then we count the number of Young tableaux <i>T</i> of two columns according to the complexity of <i>T</i>. In particular, we show that the number of Young tableaux <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(T\in Tab_{(2^k,1^{n-2k})}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>T</mi> <mo>∈</mo> <mi>T</mi> <mi>a</mi> <msub> <mi>b</mi> <mrow> <mo stretchy="false">(</mo> <msup> <mn>2</mn> <mi>k</mi> </msup> <mo>,</mo> <msup> <mn>1</mn> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mi>k</mi> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> </msub> </mrow> </math></EquationSource> </InlineEquation> such that <InlineEquation ID="IEq8"> <EquationSource Format="TEX">\(\mathcal {F}_T\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="script">F</mi> <mi>T</mi> </msub> </math></EquationSource> </InlineEquation> is a smooth is given by <Equation ID="Equ8"> <EquationSource Format="TEX">\(\begin{aligned}\left\{ \begin{array}{ll} 2\left( {\begin{array}{c}k+1\\ 4\end{array}}\right) -\left( {\begin{array}{c}k\\ 3\end{array}}\right) +\left( {\begin{array}{c}k\\ 2\end{array}}\right) +1,&amp; n=2k, \\ \frac{(k+1)(3kn-(k+2)(4k-3))}{6},&amp; n\ge 2k+1, \end{array}\right. \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mn>2</mn> <mfenced close=")" open="("> <mrow> <mtable> <mtr> <mtd> <mrow> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <mn>4</mn> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> <mo>-</mo> <mfenced close=")" open="("> <mrow> <mtable> <mtr> <mtd> <mi>k</mi> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <mn>3</mn> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> <mo>+</mo> <mfenced close=")" open="("> <mrow> <mtable> <mtr> <mtd> <mi>k</mi> </mtd> </mtr> <mtr> <mtd> <mrow> <mrow /> <mn>2</mn> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> <mo>+</mo> <mn>1</mn> <mo>,</mo> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mi>n</mi> <mo>=</mo> <mn>2</mn> <mi>k</mi> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mfrac> <mrow> <mo stretchy="false">(</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">(</mo> <mn>3</mn> <mi>k</mi> <mi>n</mi> <mo>-</mo> <mo stretchy="false">(</mo> <mi>k</mi> <mo>+</mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo stretchy="false">(</mo> <mn>4</mn> <mi>k</mi> <mo>-</mo> <mn>3</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> </mrow> <mn>6</mn> </mfrac> <mo>,</mo> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mi>n</mi> <mo>≥</mo> <mn>2</mn> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>for all <InlineEquation ID="IEq9"> <EquationSource Format="TEX">\(n\ge 2k\ge 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <mn>2</mn> <mi>k</mi> <mo>≥</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>.</p>

错误:搜索内容不能为空,请输入英文关键词
错误:关键词超出字数限制,请精简
高级检索

Counting two-column Young tableaux corresponding to smooth components of Springer fibers

  • Ronit Mansour

摘要

Let x be any nilpotent endomorphism of a vector space V of finite dimension over an algebraically closed field of arbitrary characteristic. The Springer fiber \(\mathcal {F}_x\) F x is the subset of x-stable complete flags. In the case \(x^2=0\) x 2 = 0 , the components of \(\mathcal F_x\) F x are parameterized by Young tableaux of shape \((2^k,1^{n-2k})\) ( 2 k , 1 n - 2 k ) of two columns, where \(k=\textrm{Rank}\,x\) k = Rank x . In this paper, we present an equivalent parameterization of the components of \(\mathcal F_x\) F x , and then we count the number of Young tableaux T of two columns according to the complexity of T. In particular, we show that the number of Young tableaux \(T\in Tab_{(2^k,1^{n-2k})}\) T T a b ( 2 k , 1 n - 2 k ) such that \(\mathcal {F}_T\) F T is a smooth is given by \(\begin{aligned}\left\{ \begin{array}{ll} 2\left( {\begin{array}{c}k+1\\ 4\end{array}}\right) -\left( {\begin{array}{c}k\\ 3\end{array}}\right) +\left( {\begin{array}{c}k\\ 2\end{array}}\right) +1,& n=2k, \\ \frac{(k+1)(3kn-(k+2)(4k-3))}{6},& n\ge 2k+1, \end{array}\right. \end{aligned}\) 2 k + 1 4 - k 3 + k 2 + 1 , n = 2 k , ( k + 1 ) ( 3 k n - ( k + 2 ) ( 4 k - 3 ) ) 6 , n 2 k + 1 , for all \(n\ge 2k\ge 2\) n 2 k 2 .