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Sam Nead
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TheIn general, the answer is no. ForFor an example, suppose that $K$ is the boundary of the four-simplex. Thus $K$ is a triangulation of the three-sphere. Now every closed connected oriented surface of genus $g$, denoted $M_g$, embeds in some subdivision of $K$. However, as $g$ grows, more and more subdivisions are needed. In particular, only the two-sphere, $M_g$, embeds in (the two-skeleton of) $K$.

The answer is no. For suppose that $K$ is the boundary of the four-simplex. Thus $K$ is a triangulation of the three-sphere. Now every closed connected oriented surface of genus $g$, denoted $M_g$, embeds in some subdivision of $K$. However, as $g$ grows, more and more subdivisions are needed. In particular, only the two-sphere, $M_g$, embeds in (the two-skeleton of) $K$.

In general, the answer is no. For an example, suppose that $K$ is the boundary of the four-simplex. Thus $K$ is a triangulation of the three-sphere. Now every closed connected oriented surface of genus $g$, denoted $M_g$, embeds in some subdivision of $K$. However, as $g$ grows, more and more subdivisions are needed. In particular, only the two-sphere, $M_g$, embeds in (the two-skeleton of) $K$.

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Sam Nead
  • 28.2k
  • 5
  • 72
  • 133

The answer is no. For suppose that $K$ is the boundary of the four-simplex. Thus $K$ is a triangulation of the three-sphere. Now every closed connected oriented surface of genus $g$, denoted $M_g$, embeds in some subdivision of $K$. However, as $g$ grows, more and more subdivisions are needed. In particular, only the two-sphere, $M_g$, embeds in (the two-skeleton of) $K$.