let $$D=\{(t_1,\cdots,t_d)\, |\, 0
an equivalent definition of a simplex
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$\begingroup$
general-topology
algebraic-topology
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0Write down the definitions, draw some pictures and think a bit... How do you define a simplex usually? Can you *write down* what a face is in this usual picture? If so, *which* faces don't show up here? That shouldn't be too hard. – 2011-07-17
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0@Theo Buehler: I think there are two faces excluded the face $t_1=0$ and $t_d=1$ is my understanding correct? – 2011-07-17
1 Answers
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$\Delta^{d} = \{ (x_1, \dots, x_{d+1}) \in \mathbb R^{d+1} \;|\; \sum_{k=1}^{d+1} x_k = 1 \text{ and } 0 \leq x_1, \dots, x_{d+1} \leq 1 \}$ is probably the definition of a $d$-dimensional simplex you know.
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Let $t_i := \sum_{k=1}^{i} x_k$ for $1 \leq i \leq d+1$. We now have $0 \leq t_1 \leq \dots \leq t_d \leq t_{d+1} = 1$. Thus the faces defined by $x_1=0$ and $x_{d+1} = 0$ are excluded from $D$.
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1So much for the attempt to induce palio to think... – 2011-07-17
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0Ok, I've hidden my answer, I hope this will help :D – 2011-07-17
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0Barely, I guess... – 2011-07-17
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0Well I've started writing my answer, before you got palio involved. I was disrupted for some time and didn't check for updates before submitting. I'm sorry to have disrupted your more pedagogically meaningful approach to answer his question. If you want I can still delete my answer... – 2011-07-17
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0No need to apologize or delete your question. Answering questions is what this site is for, after all :) Sorry about the temporary downvote. – 2011-07-17
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0@ Alexander Thumm: I think there are two faces excluded the face $t_1=0$ and $t_d=1$ is my understanding correct? – 2011-07-17
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0@Theo Buehler: Yes, that is what this site is for, and I interrupted your approach of doing so... @palio: Yes. Can you locate these faces in dimensions $1,2,3$ ? – 2011-07-17