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# Mazur Manifold – Class Lecture

Let G = < A | R > be a presentation of a group.

Tietze Transformations (finite versions) are ways that a group presentation can be altered without changing the group.

$$T_I$$ – add a relator that is a consequence of the other relators.

example: $$aba^{-1}b^{-1} , aba$$ implies $$a^2b$$

$$T_{II}$$ – add a new generator while simultaneously adding a relator that expresses that new generator as a word in the existing generators.

example: $$<a, b | aba^{-1}b^{-1}, aba>$$ transformed into $$<a, b, c| c= a^2b^3, aba^{-1}b^{-1} , aba >$$

$$T_{I}^{-1}, T_{II}^{-1}$$ – reverses of the above moves.

Fact: Finite presentation < A | R > and  < B | S > represent isomorphic groups if one may be obtained from the other via a finite sequence of Tietze Transformations.

Note: Also true for non-finitely presented groups if we generalize the Tietze Transformation.

Goal: To construct a compact contractible 4 – manifold, that is not homeomorphic to a $$B^4$$.

We will construct a Mazur 4-manifold. We begin with a 4 -dimensional solid torus $$S^2 \times B^3$$.

We will attach a single 2-handle which kills the fundamental group but does not yield a 4-ball. Clearly $$\partial (S^1 \times B^3 ) = S^1 \times S^2$$ Begin by creating a picture of $$S^2 \times S^2$$ by doing a single 0 – Dehn surgery on the punctured unknot.

Next, we attach a 2 – handle whose attaching sphere (i.e. circle) lies in the complement of that unknot and hence in $$S^1 \times S^2$$.

That circle is pictured.

For meridional curve in the corresponding integral Dehn surgery, choose a simple closed curve that stays parallel to $$\Gamma$$ in the given picture.

$$\pi_1 (S^3 – \Gamma \cup \Theta ) = < x_1 , … , x_9 | r_1 , … , r_9 >$$

$$\pi_1(S^1 \times S^2 – \Gamma ) = < x_1 , …, x_9 | r_1, … r_9 , x_5 x_2^{-1}x_1^{-1} >$$

$$\pi_1 ( \partial M^4 ) = <x_1 , … , x_9 | r_1, …, r_9, x_5x_2^{-1}x_1^{-1}, x_7^{-1} x_5^{-1} x_7 x_3^{-1} x_2^{-1} x_7^{-1} >$$

Perform Tietze Transformations

Let $$\beta = x_7 , \lambda = x_2 , \alpha = \beta \lambda$$

Perform Tietze Transformations

$$\pi_1 ( \partial M^4 ) \equiv < \alpha , \beta | \beta^5 = \alpha^7 , \beta^4 = \alpha^2 \beta \alpha^2 >$$

Goal: To show that this group is non-trivial.

First, we will introduce an additional relator $$\beta^5 = 1$$

The resulting group is the quotient group $$\pi_1 (\partial M^4 ) / << \beta^5 >>$$. If it is non-trivial then so is $$\pi_1 ( \partial M^4 )$$.

We now have $$< \alpha, \beta | \alpha^7 , \beta^5, \beta^4 = \alpha^2 \beta \alpha^2 >$$ ## By Ashani Dasgupta

Pursuing Ph.D. in Geometric Group Theory at University of Wisconsin, Milwaukee