In [Arn1] Arnold likes to use figures to illustrate his ideas. This saves a lot
of words in explaining ecological phenomena. [Arn1, p.30, Fig.17 & Fig.18] list
the interesting possibilities for oscillation. We also like to see the
metamorphosis of Lamercy diagrams and the phase portrait [Arn1, p.46, Fig. 36;
p.47, Fig.37] caused by a small perturbation. In contrast, [Pon, pp.220-236,
§28] pays too much attention to
detailed analysis. The more details we investigate, the more things we find unclear.
On the one hand, we can never close our topic; on the other hand, details blur
the big picture.
The constructions with a picture in mind.
[War, pp.132-133, Exercise 7 & Exercise 8] describe the picture of a Riemann
surface in an insightful way [Lev, p.73, Fig.4-4 or p.76, Fig.4.5]. Too many
details in a description may result in losing the big picture. Exercise 7 (a)
(For details, see [Mas, p.156, Theorem 5.1]) & (c) sketch the construction of a
lift in terms of curves. Exercise 7(b) sketches (For details, see [Mas, p.175,,
l.16-p.177, l.9]) the construction of the topology of the Riemann surface. [War,
p.133, Exercise 8] (For details, see [Mas, p.181, l.3-p.183, l.3]) & [War, p.99,
l.13-l.19] sketch the construction of its manifold.
The big picture emphasizes one story and the interconnection of
What problem do we try to solve? What is the role played by a step in
the whole process?
In discussing Gauss' elimination
algorithm, we summarize the process into three steps [Rut, p.19, l.13-l.!7].
The first two steps (triangular factorization and forward substitution) are
by-products. The third step (back substitution) is the main step to produce the
solution. Therefore, it is natural to put the three steps into one story.
When we must treat these three steps separately, we should point out the reason
why we prefer doing so on this occasion [Rut, p.21, l.-8].
If one fails to follow the natural setting, timing, and order, he can
never get things straight even if he spends tremendous effort (Compare [Rut,
pp.16-19, §2.2] (Good!) with [Baz,
p.559, l.1-l.25] (bad!)).
The big picture focuses on the origin and inspires an effective
Example (Triangular factorization). The scheme of uppertriangularizing a
matrix and the scheme of retaining computing information work together to
produce the two desired factors simultaneously [Rut, p.16, l.-19-p.18,
The big picture focuses on the idea's
continuity and the typical case.
In [Baz, p. 559, l.1-l.25], Bazaraa interrupts the flow
of ideas with his side considerations of permutations [Baz, p.559, l.5-l.16], which should
be treated as exceptional cases. His improper interruption makes a simple idea
Our ultimate goal of understanding a theorem is to be able to see
its geometric picture.
Example. Interpreting [Baz, p.47, Gordan's
Theorem] in a way similar to [Baz, p.47, Fig. 2.10].
The proof of [Bir, p.268, Theorem 3] lacks a picture. Consequently, it seems
that when making trivial points, Birhoff engaging in discussion that provide great detail;
when presenting points essential to reader's understanding, his descriptions are
vague and sketchy. His lack of communication skills in geometry greatly impedes understanding. The following
may help his readers to visualize his argument.
In [Bir, p.268, l.16], "the number of zeros of sin
q1(x) for a< x <b is greater than the number of zeros of
q(x)" means "the number of times that
q1 crosses P1u1'-axis
is greater than the number of times that q crosses
In [Bir, p.268, l.-15], "between any two zeros of u(x)" means "
q goes through at least the length of
Other comments related to pictures.
In [Bir, p.268, l.-1], "A solution of (20) has exactly one maximum or minimum between successive zeros" means
"As q increase from np
to (n+1)p, only q = (n+½)p
satisfies cos q = 0".
In [Bir, p.270,
l.12], "To prove this result, it suffices to prove that
q(x,l) is an increasing function of
l" should be changed to "It suffices to prove
that q(x,l) is an increasing function of
l. This is because xn(l)
is a continuous function of l".
[Coh, pp.74-77, Complement HI,
discusses the Potential Well of Finite Depth in great detail. However, in order
to see its big picture, we must view the topic as the boundary value problem
of the Schrödinger equation [Bir, p.290,
When a physicist tries to prove the interchangeability of the order of a
summation and an integral, he immediately conjures up the tricks of uniform
convergence [Guo, p.357, l.-8] and
differentiation under the integral sign [Guo, p.357,l.-7].
This is like someone falling into a lake and desperately trying to grab
anything as a
life safer. In [Guo, p.357, l.-10-p.358, l.3], Guo
performs a lot
of actions, but I cannot see they contribute to the proof. The key idea behind the
proof is the Fubini Theorem [Ru2, p.150, Theorem 7.8]. All we need to do is
to check if the condition [Ru2, p.150, (3)] is satisfied.
Computer-generated figures vs. hand-drawn figures
Computer-generated figures usually look better than
hand-drawn figures. If possible, people prefer to use computer-generated
figures. However, a figure-drawing computer software program may have
limited features. If two figures
are both generated by a computer, they usually repel each other. This can
make the meaning of the figures ambiguous. For example, in [Mun00, p.380,
Fig. 61.3], one wonders which curve f and g map into, the loop containing x0
or the curve A1. If arrowheads of f and g
could touch the loop, then it would be clear that f and g map into the loop.
Therefore, it would be great if someone can combine the two computer-generated figures into
one in a very precise manner.
How we analyze figures [Ahl, p.98, Fig, 13; p.99, Fig. 14] In order to visualize the Riemann surface for w = cos z, one needs some imagination
in addition to rigorous analysis.
Riemann Surfaces In order to impose a coordinate system
on [Ahl, p.99, Fig. 14], we must label the vertical lines as follows: The
first line from the left is x = 0; the second line from the left is x =
p, and so on. The horizontal
line is y = 0. Then we must label the sheets in [Ahl, p.98, Fig. 13] as follows:
The first sheet from the top is k = -1;
the second sheet from the top is k = 0, and so on. Since x = 2np
corresponds to the positive cut, on sheet 2n we may focus on the branch point
w=1. Since x = (2n+1)p
corresponds to the negative cut, on sheet 2n+1 we may focus on the branch point
w = -1. We
may draw a curve with the positive sense around x = 0
so that its image path around w = 1 on sheet -1
of the Riemann surface for w = cos z will be shaped like a closed path
around w = 0 on the Riemann surface for w = z1/2 [Lev, p.76, Fig. 4-5].
Indeed, we may start from a point w0 that is
to the right of w = 1 and on sheet 0. For the first round, the image path
decreases one level to sheet -1;
for the second round, the image path increases one level to sheet 0 and returns to
the starting point w0. This explains the
situation around the top right branch point w = 1 in [Ahl, p.98, Fig. 13]. The
situations around other branch points in [Ahl, p.98, Fig. 13] can be
Example 1. [(z+1)(z-1)-1]1/2 maps onto Im w > 0
of the domain obtained from the z-plane by taking two cuts,
defined by -¥ £ x £
1 and 1 £ x < +¥,
along the real axis [Fuc, pp.94-95, Example 2]. Ahlfors' description
given in [Ahl, p.243, l.1] is incorrect even though he and Sario wrote a
book called Riemann Surfaces. Thus, one who understands a
theory thoroughly may not fully understand each example in the
development of the theory.
Example 2. The Riemann surface for the Bessel function of the first kind:
[Wat, p.75, (1) & (2)] follow from [Wat, p.40, (8)].
Branch points and cuts
A point where all the sheets of the Riemann surface w =
f(z) meet is called a branch point. A branch point can be finite or
¥. In the former
case, we may assume w = 0; in the latter case, we let w =
¥. For w = log z,
the branch point is w = ¥
[Lev, p.73, Fig. 4-4]; for w = z1/2, the
branch point is w = 0 [Lev, p.76, Fig. 4-5]; for w = z-1/2,
the branch point is w = ¥.
[Gon1, p.400, Fig. 5.10]. A cut can be any simple arc on one sheet extending from a
branch point to ¥
[Ahl, p.98, l.12-l.13].
A branch cut serves as a reminder that a contour of integration enters a
different sheet of the Riemann surface when it crosses the cut [Hob,
p.192, Fig. (a) and Fig. (b)]. One usually chooses a simple and
convenient dividing line between one sheet and the next as one's
cut [Hob, p.184, l.-19-l.-8;
Wat1, p.312, l.4-l.8].
A contour integral can be used to define a single-valued function if we
do not allow the contour to cross the
branch cut [Hob, p.192, l.15-l.-1]. Example. 1 and ¥ are
the branch points of the hypergeometric function F(a, b; g; z) [Wat1, p.291, l.3-l.5; p.289; l.3-l.4].
The explanation given in [Guo, p.151, l.-2-p.152, l.-12] is poor.
Branch points vs. images of poles [Gon1, p.397, l.-2-l.-1] The
denominators of both functions w =1/z and w =1/(z1/2)
are 0 when z = 0. The former function w(z) is single-valued in the neighborhood
of z = 0, while the later function w(z)
is multiple-valued in the neighborhood of z = 0.
Phases and integrals on Riemann surfaces
Only through the tool of Riemann surfaces may we clarify the confusion about the phase problem.
Example 1 [Wat1, p.257, l.11-l.13]. On the Riemann surface for log (t -
1), any point must be assigned a phase; otherwise its power cannot be
defined. For example, if t is a point on the contour given on [Wat1, p.257,
t - 1 = |t -
1|eiq; 1 - t = epi (t - 1).
Case 1: t is on path a. t - 1 = |t -
1|e-pi = (1 - t) e-pi.
(1 - t)a-1 = [epi (t -
1)]a-1 = (epi )a-1(1 -
Case 2: t is on path b. t - 1 = |t -
1|epi = (1 - t) epi.
(1 - t)a-1 = [epi (t -
1)]a-1 = (epi )a-1(1 -
Example 2 [Wat, p.166, l.8]. In [Wat, p.165, Fig.4], let arg (t -
1)|t = A = 0; then arg (t -
1)|t = B = p (Consider the Riemann surface of log (t - 1)).
Example 3 [Wat, p.166, l.11-l.17]. Let the contour be t = -1 + r(q) eiq, where r(q+2p) = r(q).
If q increase 2p [Wat, p.166, l.11-l.12] (i.e., going up a level on the sheets of the Riemann surface of log (t+1)), then q should be replaced by q - 2p.
Then (t+1)n-1/2 should be replaced by e-2p(n-1/2)i (t+1)n-1/2.
e-2p(-1/2)i = epi is used to reverse the direction of the contour, i.e., - ò-1+¥i(-1+) = ò-1+¥i(-1-) .
Combining e-2pni with enpi given in [Wat, p.166, l.5], we have e-npi.
The integral ò-1+¥i(-1-) given in [Wat, p.166, l.4] originates from [Wat, p.165, Fig.4], while the integral ò-1+¥i(-1-) given in
[Wat, p.166, l.14] originates from [Wat, p.165, Fig.5]. The integration paths
for the two integrals can be in different sheets of the Riemann surface for log (t - 1). In order to make sure that the two integrals represent the same value, we must assign the same value p for the phase of t + 1 at
t = B [Wat, p.166, l.11-l.12]. Actually, one can choose any point on the
integration contour to assign a phase [Wat, p.166, l.11-l.12; Guo, p.370,
l.-4] as long as the contour is on the same sheet of Riemann surface for log
(t - 1).
Example 4 (Derive the formula given on [Wat, p.164, l.-7-l.-3] from [Wat, p.164, (2)]).
Case ò0¥i: We consider this case first because, its phase can be easily determined.
t + 1 =|t+1| eiq, where 0 £ q £ p/2.
t - 1 =|t-1| ei (p-q)
t2 - 1 = |t2 - 1|epi = (1 - t2)epi .
Case ò10: the phase of t + 1 is 0; the phase of t - 1 is p.
Case ò-11: the phase of t + 1 is 0; the phase of t - 1 is -p.
Case ò0-1: the phase of t + 1 is -2p; the phase of t - 1 is -p.
Case ò¥i0: the phase of t + 1 is -2p + q, where
0 £ q £ p/2; the phase of t - 1 is -p - q.
|t2 - 1| = 1 - t2.
Ahlfors' use of the reflection principle to
extend the domain of w = sn z [Ahl, p.231, l.-5-p.233, l.2]
yields half the result with twice the effort.
In contrast, the figure of w = sn z given in [Gon1, p.424, Fig. 5.19]
provides a bird's-eye view which facilitates understanding of everything that Ahlfors describes in [Ahl, p.231, l.-5-p.233, l.2]. The figure
w = sn z given in [Gon1, p.424, Fig. 5.19] is derived from the theory of
The passage given in [Mun00, p.363, l.4] serves to demonstrate the
important features of [Mun00, p.364, Fig. 58.3] rather than give the
detailed proof of [Mun00, p.363, Lemma 58.4]. These features easily lead to
the proof of [Mun00, p.363, Lemma 58.4]. However, one who understands the details of the proof
may not necessarily recognize these important features in the figure.
The point of a picture must be clearly explained [Wat1, p.307, l.16-l.-9
; p.312, l.6-l.7].
The two contours given in [Wat1, p.307] meets at point A but lie on different
sheets of a Riemann surface, so the values of the integral given in [Wat1,
p.307, l.7] along these two contours are different. From the
viewpoint of homotopy, the explanation for [Hob, p.192, Fig. (a) and Fig. (b)] is
better than that for [Wat1, p.307, Fig.].