| Page |
Position |
Replace |
by |
Thanks go to |
| 15 |
Eq. (1.33) |
(P⇒Q) ∧ (Q⇒R) (P⇒R) |
(P⇒Q) ∧ (Q⇒R)
(P⇒R) |
Viktor Skorniakov, Vilnius University |
| 22 |
Table in Eq. (2.17), 2nd row |
T T T T T |
T F F F F |
Viktor Skorniakov, Vilnius University |
| 25 |
Eq. (2.26) |
f(G) = {y ∈ Y | ∃ x: f(x)=y} |
f(G) = {y ∈ Y | ∃ x f(x)=y} |
Viktor Skorniakov, Vilnius University |
| 28 |
Example 3 |
Matrix multiplication is a binary operation on the
said set. Matrix addition is a binary operation on the set
of all n×n matrices. |
Viktor Skorniakov, Vilnius University |
| 39 |
Eq. (3.4) |
The expression on the right
side should be placed under a square root sign. |
Viktor Skorniakov, Vilnius University |
| 51 |
Next to last sentence in the proof of Theorem 3.10 |
Now, one can write an arbitrary combination of a
and b as ax + by = |
Now, one can write an arbitrary combination of a
and b as ax + by = |
Viktor Skorniakov, Vilnius University |
| 69 |
2nd line after Eq. (4.7) |
something that is more commonly denoted by |
something that is more commonly denoted by |
Viktor Skorniakov, Vilnius University |
| 70 |
Example 2 |
The example given is a valid
example of a linear map, which is an
element in the dual space. It does not
describe the full dual space. There are
other maps, of course.
| Charanjit Singh Aulakh, IISER Mohali |
| 73 |
Eq. (4.22), right side |
xi2 |
|xi|p |
|
| 98 |
Eq. (5.60) |
e'(j) = ∑k
(S-1)ik e(i) |
e'(k) = ∑i
(S-1)ik e(i) |
Charanjit Singh Aulakh, IISER Mohali |
| 104 |
After Eq. (5.98) |
The proof of the Cayley-Hamilton theorem is not
trivial. The result is an operator relation, as given in
Eq. (5.98), and cannot be inferred from Eq. (5.95) which is a
determinant equation. The proof mentioned in the book is often
called the bogus proof of the theorem. Just ignore the
"proof". The theorem has been rightly applied in what follows
immediately, and elsewhere.
|
Charanjit Singh Aulakh, IISER Mohali |
| 124 |
Eq. (5.222) |
U = ∑i ,
V = ∑i .
|
U = ∑i ,
V = ∑i .
|
Charanjit Singh Aulakh, IISER Mohali |
| 128 |
2nd line after Eq. (6.5) |
If the Wronskian vanishes, the
functions are linearly |
If the Wronskian vanishes, the
functions are linearly |
Joydeep Sarkar, Chennai Mathematical Institute |
| 174 |
after Eq. (7.63) |
hi−1 also commutes
with |
hi−1 also commutes
with |
Sayan Chakrabarti, IIT Guwahati |
| 176 |
Th. 7.16 |
The theorem, as stated, is wrong. It is best to
disregard the statement as well as the proof. |
Charanjit Singh Aulakh, IISER Mohali |
| 185 |
Eq. 7.100 |
img(fk) = ker(fk+1) |
img(fk-1) = ker(fk) |
Charanjit Singh Aulakh, IISER Mohali |
| 207 |
2nd line after Eq. (8.77) |
imply b=a |
imply b=a |
|
| 222 |
last line |
one 2-cycle and three 1-cycles |
one 2-cycle and two 1-cycles |
Inayat R. Bhat |
| 229 |
Eq. (9.5), 1st line |
U†Ai†AiU† |
U†Ai†AiU |
Trinesh Sana, Calcutta University |
| 239 |
Eq. (9.64) |
The right side of the equation should have an additional
factor of 1/|G|. |
Inayat R. Bhat |
| 263 |
row for A3 in Eq. (9.149) |
|
|
Inayat R. Bhat |
| 3rd line of text after Eq. (9.149) |
the elements of the first three classes form a
subgroup of rotations |
the elements of the first, third and fourth classes
form a subgroup of rotations |
Inayat R. Bhat |
| rows for A3 and A4 in Eq. (9.150) |
|
|
Inayat R. Bhat |
| 265 |
just before Eq. (9.155) |
classes of S3 |
classes of S4 |
Luis Odin Estrada Ramos, UNAM, Mexico |
| 270 |
Eq. (9.178) |
The places of A3
and A4 should be interchanged. |
Inayat R. Bhat |
| 296 |
End of 6th line after Eq. (10.50) |
and α3
= 1 + 1 + 2 = 4. |
and α6
= 1 + 1 + 2 = 4. |
|
| 301, 302 |
Eqs. (10.73), (10.74) |
The arguments of all sin and
cos functions should be 2πk/M;
there should not be any factor
of i. |
Charanjit Singh Aulakh, IISER Mohali |
| 316 |
Heading of Section 11.3 |
The group denoted
by SL(2,Z) in the
heading has been
called SL(2Z) within
the text. The notations with and without
the comma mean the same thing. |
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 334 |
1st paragraph |
There are two different uses of the word "semisimple", as applied to
algebras. In this paragraph, the word "semisimple" has been used in
the exclusive sense, i.e., these are algebras which are neither
simple, nor do they have any abelian ideal. Beginning from the next
paragraph, the same word has been used in an inclusive sense, i.e.,
by simply demanding that they do not have any abelian ideal. In the
latter sense, simple algebras also fall in the class of semisimple
algebras. One should interpret Theorem 12.2 in this way, as well
as Theorems 12.7 and 12.8 later in the chapter.
Alternatively, one can use the inclusive definition throughout by
making the following changes in this paragraph:
| Augniva Ray, Saha Institute of Nuclear
Physics, Calcutta |
| If an algebra is non-simple but not in this trivial way, i.e.,
not by the presence of any U(1) generator that commutes with every
generator, then it is called a semisimple algebra. We will
see in Chapter 15 that SO(4), the algebra of the group of 4×4
orthogonal matrices, is semisimple. |
If an algebra is not non-simple in this trivial way, i.e.,
if an algebra does not have any U(1) ideal, then it is called
a semisimple algebra. The definition therefore includes
simple algebras which do not have any non-trivial ideal at all, as
well as other algebras which have non-trival ideals which are all
non-abelian. We will see in Chapter 15 that SO(4), the algebra of
the group of 4×4 orthogonal matrices, is of this last kind,
i.e., is semisimple but not simple. |
| 337 |
midway in the 2nd paragraph |
But the total exponent must be Hermitian so that R(ξ)
is unitary, which means that the exponent must also contain the
Hermitian conjugate of −iξX, which
is +iξ*X†. |
But the total exponent must be of the form of the imaginary
unit i times a Hermitian operator so that R(ξ) is
unitary, which means that the terms involving ξ in the exponent
must be of the form −i(ξX +
ξ*X†). |
Charanjit Singh Aulakh, IISER Mohali |
| 338 |
Eq. (12.35), right side |
KI |
aIKI |
Charanjit Singh Aulakh, IISER Mohali |
| 339 |
Eq. (12.45), second line |
[A,[B,C]+] +
[B,[C,A]+]+ −
[C,[A,B]+]+ = 0 |
[A,[B,C]+] +
[B,[C,A]]+ −
[C,[A,B]]+ = 0 |
Trinesh Sana, Calcutta University |
| 350 |
Eq. (12.95), last term on the right side |
Ta†(2) ⊗
Tb(1) |
Tb(1) ⊗
Ta†(2) |
Charanjit Singh Aulakh, IISER Mohali |
| 351 |
Eq. (12.98), 3rd term on the left side |
(2⁄d(ad))
tr (Ta†(1))
tr (Ta(2)) |
(1⁄d(ad)) [
tr (Ta†(1))
tr (Ta(2)) +
tr (Ta(1))
tr (Ta†(2))
] |
Charanjit Singh Aulakh, IISER Mohali |
| 356 |
Section 12.10.4 |
The entire argument, leading to Eq. (12.124), can be taken as an
argument for the normalization constant of a representation of a
direct product group, replacing C2 by K.
For the Casimir invariant, the factors of the dimensions of the
representations should be omitted in all three equations of this
section. It means that the final formula should be
C2(R,R') =
C2(R) +
C2(R'),
with appropriate changes in the two earlier equations. The dimensions
are important only to the extent that the unit matrix
is d(R)d(R')-dimensional in these earlier
equations.
| Charanjit Singh Aulakh, IISER Mohali |
| 383 |
Eq. (13.162), under the square root sign |
s(s + 1) ± m(m
− 1) |
s(s + 1) − m(m
± 1) |
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 386 |
Table 13.2, first table
|
In the column marked m1, the entry
for the second row should be −½, not ½. |
Augniva Ray, Asia-Pacific Center for Theoretical Physics |
| 422 |
2nd line after Eq. (15.16) |
if V is |
if M is |
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 426 |
Eq. (15.31) |
There should not be a minus
sign on the right side of this equation. |
Trinesh Sana, Calcutta University |
| 439 |
line after Eq. (15.80) |
Recalling that C−1 =
C−1 |
Recalling that (C*C)−1 =
C−1 CT |
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 474 |
first line of text |
denote by Λ a matrix |
denote by Λ a matrix |
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 482 |
2nd line of Eq. (17.57), left side |
t' − z' |
ct' − z' |
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 498 |
Eq. (17.143) |
The right sides of
Eqs. (17.143b) and (17.143d) should have
overall negative signs. |
Amitabha Lahiri, S N Bose Centre for
Basic Sciences |
| 577 |
Eq. (19.148) |
xi → x'i
+ εi(x)
|
xi → x'i = xi
+ εi(x)
|
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 587 |
Eq. (19.211) |
In the expression
for h(t), the sum should be over n.
|
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 589 |
Eq. (19.221), the first set of nested brackets |
[Tam,
[Tbn,
Tpc]]
|
[Tam,
[Tbn,
Tcp]]
|
Augniva Ray, Saha Institute of Nuclear Physics, Calcutta |
| 589 |
Eq. (19.222) |
fabd δdc 𝜙(m + n) +
fbcd δda 𝜙(n + p)
fcad δdb 𝜙(p + m)
|
fabd δdc 𝜙(m + n) +
fbcd δda 𝜙(n +
p) +
fcad δdb 𝜙(p + m)
|
|
| 604 |
3rd line before Eq. (20.19) |
For the open ball G in Y, we
find |
For the open set G in Y, we
find |
Viktor Skorniakov, Vilnius University |
| 616 |
1st line of text |
the rectangle marked 'B' is (150,185)× |
the rectangle marked 'B' is (120,185)× |
Viktor Skorniakov, Vilnius University |
| 617 |
Eq. (21.23) |
{G ∩ S ∀G ∈ T} |
{G ∩ S ∣ ∀G ∈ T} |
Viktor Skorniakov, Vilnius University |
| 619 |
Eq. (21.31) |
U∈Y |
U⊂Y |
Viktor Skorniakov, Vilnius University |
| 620 |
Eq. (21.38) |
Subset U∈Y |
Subset U⊂Y |
Viktor Skorniakov, Vilnius University |
| 626 |
Definition (21.18) |
there exist non-empty subsets |
there exist non-empty open subsets |
Viktor Skorniakov, Vilnius University |
| 637 |
Eq. (22.23) |
In the last line of the defintion of H(s,t),
the function should be f̅ instead of f. |
Viktor Skorniakov, Vilnius University |
| 655 |
Definition (23.2) |
is a map |
is a continuous map |
Viktor Skorniakov, Vilnius University |
| 661 |
Eqs. (23.25) to (23.27) |
Summation over the index i should range from 0
to d. |
Joydeep Sarkar, Channai Mathematical Institute |
| 662 |
Text before Eq. (23.30) |
The most general element of C1(X) is given
already in Eq. (23.26). |
The most general element of C1(X) is given
already in Eq. (23.25). |
Viktor Skorniakov, Vilnius University |
| 664 |
Eqs. (23.43) & (23.45) |
The right side of Eq. (23.43) is really the image of
∂2 and not its kernel. The kernel can be obtained by
equating the right side of Eq. (23.35) to zero, which
gives p=0. Ths means that the kernel is trivial: ker
(∂2) = {0}. Using Eq. (23.44) now, one
obtains H2(D2) = {0}/{0} =
{0}.
|
Joydeep Sarkar, Channai Mathematical Institute |
| 684 |
Answer of Ex. 14.8 |
8, 10, 10, 6 |
8, 15, 15, 6 |
Augniva Ray, Saha Institute of Nuclear
Physics, Calcutta |
| |
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