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\large{2014 Fall Math 5041 \quad Assignment 2. \qquad Due: Oct 20, 2014}
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\large{Name: }
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1.
a) Let $\mathbb{R} \to GL(2, \mathbb{R})$ be a smooth one-parameter family of invertible $2 \times 2$ matrices. Show that
$$
	\frac{d \left( \det A \right) }{dt} = \left( \det A \right)  \mathtt{tr} \left(A^{-1} \frac{d A}{dt}\right).
$$

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b) Show that the special linear group
$$
	SL(2, \mathbb{R}) = \{A \in \mathtt{Mat}_{2\times 2} ~|~ \det A = 1\}
$$
is a Lie group.

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Remark: This works in higher dimensions as well. A more conceptual way is as follows. The determinant map 
$$
	\det: GL(n, \mathbb{R}) \to \mathbb{R}^{*}
$$
is a Lie group homomorphism, which is a smooth surjective map of constant rank. Therefore, $1$ is a regular value of $\det$ and
$$
	SL(n, \mathbb{R}) = \ker (\det) = {\det}^{-1} (1)
$$
is a submanifold of $GL(n, \mathbb{R})$.
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2. (cf. Problem 4-6 [Lee pp. 96])
Let $M$ be a non-empty smooth compact manifold (without boundary). Show that there is no smooth submersion $f: M \to \mathbb{R}^k$ for $k > 0$.

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3. Let $K$ and $L$ be regular submanifolds of $M$. We say $K$ and $L$ intersect cleanly if $K \cap L$ is also a regular submanifold of $M$ and for every point $p \in K \cap L$, we have
$$
	T_p (K \cap L) = T_p K \cap T_p L.
$$

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a) Show that transversal intersection implies clean intersection.

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b) Show that clearn intersection does not imply transversal intersection.

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4. A smooth map $f: M \to N$ is called a proper map if and only if $f^{-1}(K) \subset M$ is compact for very compact subset $K \subset N$. Prove that an injective proper immersion is an embedding.

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5. Let $f : M \to M$ be a smooth map. For a fixed point $p$ for $f$, i.e. $f(p) = p$, we say $p$ is a Lefschetz fixed point when the derivative map $Df : T_pM \to T_pM$ does not have $+1$ as an eigenvalue.
Show that if M is compact and all fixed points for $f$ are Lefschetz, then there are only finitely
many fixed points for f.

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6. Construct a smooth map $f: \mathbb{R} \to \mathbb{R}$ such that the set of critical values is dense.

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Remark: A dense subset of $\mathbb{R}$ could have measure zero, e.g. the set of rational numbers, so it does not contradicts Sard's theorem.
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