The degree condition in Llarull’s theorem on scalar curvature rigidity

Llarull’s scalar curvature rigidity theorem states that a <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mn>1</mml:mn> </mml:math> -Lipschitz map <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>f</mml:mi> <mml:mo lspace="0pt">:</mml:mo> <mml:mi>M</mml:mi> <mml:mo>→</mml:mo> <mml:msup> <mml:mi>𝕊</mml:mi> <mml:mi>n</mml:mi> </mml:msup> </mml:mrow> </mml:math> from a closed connected Riemannian spin manifold <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mi>M</mml:mi> </mml:math> with scalar curvature <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mo form="prefix">scal</mml:mo> <mml:mo>≥</mml:mo> <mml:mi>n</mml:mi> <mml:mo>(</mml:mo> <mml:mi>n</mml:mi> <mml:mo>-</mml:mo> <mml:mn>1</mml:mn> <mml:mo>)</mml:mo> </mml:mrow> </mml:math> to the standard sphere <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:msup> <mml:mi>𝕊</mml:mi> <mml:mi>n</mml:mi> </mml:msup> </mml:math> is an isometry if the degree of <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mi>f</mml:mi> </mml:math> is nonzero. We investigate if one can replace the condition <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mo form="prefix">deg</mml:mo> <mml:mo>(</mml:mo> <mml:mi>f</mml:mi> <mml:mo>)</mml:mo> <mml:mo>≠</mml:mo> <mml:mn>0</mml:mn> </mml:mrow> </mml:math> by the weaker condition that <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mi>f</mml:mi> </mml:math> is surjective. The answer turns out to be “no” for <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>n</mml:mi> <mml:mo>≥</mml:mo> <mml:mn>3</mml:mn> </mml:mrow> </mml:math> but “yes” for <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>n</mml:mi> <mml:mo>=</mml:mo> <mml:mn>2</mml:mn> </mml:mrow> </mml:math> . If we replace the scalar curvature by Ricci curvature, the answer is “yes”in all dimensions.