Einstein summation convention: Question bank — Hinglish deep dive

4.10.9 · D5 · Maths › Advanced Topics (Elite Level) › Einstein summation convention

Poore note mein, Einstein summation convention ke yeh do load-bearing rules yaad rakho:

ek index jo ek term mein bilkul do baar aata hai woh summed hota hai (ek dummy index, aur result mein se gayab ho jaata hai);
ek index jo bilkul ek baar aata hai woh free hai, sum nahi hota, aur dono sides pe identically appear karna chahiye.

Hum helpers Kronecker delta $δ_{ij}$ aur Levi-Civita symbol $ε_{ij k}$ bhi use karte hain.

Neeche ke do pictures poora mental model hain. Pehli dikhati hai ki ek dummy index kaisa loop mein band ho jaata hai (contraction = ek sum jo vanish ho jaata hai), jabki ek free index ek khuli taar ki tarah latakta hai (ek survivor jise tum carry karte ho). Doosri ek matrix–vector product ke through actual index flow trace karti hai.

True ya false — justify karo

Expression

a_{i} b_{i}

ek scalar hai.

True. Index

i

do baar appear karta hai, isliye sum ho jaata hai, zero free indices bachte hain — aur zero free indices matlab ek akela number (dekho Dot product).

Expression

A_{ij} x_{j}

ek scalar hai.

False. Sirf

j

repeated hai (summed);

i

ek baar appear karta hai, isliye yeh ek free index hai. Ek free index ka matlab result ek vector hai —

i

ki har value ke liye ek number.

a_{i} b_{j}

(koi repeated index nahi) ka convention mein well-defined meaning hai.

True, lekin yeh sum nahi hai — yeh outer product hai, ek order-2 object with two free indices

i

aur

j

(products

a_{i} b_{j}

n \times n

grid).

δ_{ii} = 1

teen dimensions mein.

False. Yahan

i

repeated hai, isliye sum hota hai:

δ_{ii} = δ_{11} + δ_{22} + δ_{33} = 3

n

dimensions mein

δ_{ii} = n

(jahan

n

dimension hai). "Identity diagonal par

1

hai" wali fact ek single entry ke baare mein hai, summed trace ke nahi.

ε_{ij k} a_{j} a_{k}

a

ka apne aap se cross product hai, aur yeh

0

ke barabar hai.

True. Yeh

(a \times a)_{i}

hai, jo zero hai. Sirf index wali reason: dummy names

j \leftrightarrow k

swap karne se

a_{j} a_{k}

unchanged rehta hai lekin

ε

ka sign flip ho jaata hai, isliye sum apne negative ke barabar hai — hence

0

Equation

y_{k} = A_{ij} x_{j}

valid notation hai.

False. Left side ka free index

k

hai; right side ka free index

i

hai. Free indices dono sides par identical hone chahiye, isliye yeh meaningless hai — ek built-in error detector ne pakad liya.

Tum

a_{i} b_{i}

mein

i

k

se rename kar sakte ho bina value change kiye.

True. Ek dummy index sirf summation variable ka naam hai:

\sum_{i} a_{i} b_{i} = \sum_{k} a_{k} b_{k}

. Isliye

a_{i} b_{i} = a_{k} b_{k}

Tum

y_{i} = A_{ij} x_{j}

mein free index

i

ko sirf right side par

k

se rename kar sakte ho.

False. Ek free index koi private summation variable nahi — yeh label karta hai ki tum kaun sa component mean kar rahe ho, aur yeh puri equation mein match karna chahiye. Ise dono sides par rename karo ya bilkul mat karo.

ε_{ij k} ε_{ij k} = 6

3D mein.

True. Teeno indices summed hain. Yeh non-zero permutation triples count karta hai: unke

6

hain

(3!)

, har ek

(\pm 1)^{2} = 1

contribute karta hai, isliye total

6

hai.

δ_{ij} ε_{ij k} = 0

True.

δ

force karta hai

i = j

, lekin

ε

vanish hota hai jab bhi do indices equal hon. Isliye har surviving term zero hai.

Error dhundo

"Teen vectors combine karne" ke liye

a_{i} b_{i} c_{i}

likhna kya galat hai?

Index

i

teen baar appear karta hai, jo forbidden hai: ek repeated index matlab ek pairwise sum, aur teen copies pairing ko ambiguous chhod deti hain. Agar tum genuinely

\sum_{i} a_{i} b_{i} c_{i}

chahte ho, toh explicit

\sum

rakho.

Ek student

a_{i} b_{i} c_{i} d_{i}

likhta hai "do alag dot products" ke liye. Yeh kyun tuta hua hai, aur tum ise kaise fix karte ho?

Jaisa likha hai,

i

ek term mein chaar baar appear karta hai — illegal. Dummy indices ek term mein local hote hain lekin reusable nahi hote within it; ek pair rename karo:

a_{i} b_{i} c_{k} d_{k}

A_{ij} B_{ij}

matrix product

A B

hai" kyun galat hai?

i

aur

j

dono repeated hain, isliye dono summed hain — result ke zero free indices hain, ek scalar (double contraction / Frobenius inner product). Asli Matrix multiplication ek single middle index share karta hai:

C_{ik} = A_{ij} B_{j k}

Ek student claim karta hai

δ_{ij} δ_{ij} = δ_{ii} = n

. Slip pakdo.

Pehla step secretly

i

ko chaar baar reuse karta hai. Dhyan se karo:

δ_{ij} δ_{ij} = \sum_{i, j} δ_{ij}^{2} = n

(sirf

i = j

diagonal terms

1

hain; yahan

n

dimension hai). Answer

n

hai, lekin "

= δ_{ii}

" ek written expression ke roop mein illegal notation hai.

(a \times b)_{i} = ε_{j k} a_{j} b_{k}

mein kya galat hai?

3D mein Levi-Civita symbol ko teen indices chahiye,

ε_{ij k}

. Free

i

drop karna cross product ki vector nature bhi destroy kar deta hai — correct form hai

ε_{ij k} a_{j} b_{k}

Koi likhta hai

δ_{ij} a_{j} = a_{j}

. Right side kyun galat hai?

δ

ka substitution rule summed index ko free wale se rename karta hai:

δ_{ij} a_{j} = a_{i}

a_{j}

nahi.

a_{j}

likhne se right par ek free

j

bachta hai lekin left ka free index

i

hai — ek mismatch.

Curved space ya special relativity mein

a_{i} b_{i}

likhna, dono indices "downstairs" ke saath, kyun dangerous hai?

Non-Euclidean spaces mein convention ek upper index ko ek lower ke saath pair karta hai:

a^{i} b_{i}

. Do lower indices ka sum,

a_{i} b_{i}

, coordinate-independent nahi hai — tum pehle ek index ko metric

g_{ij}

ke saath lower karte ho:

a_{i} = g_{ij} a^{j}

. Flat Euclidean space mein Cartesian axes ke saath

g_{ij} = δ_{ij}

, isliye up/down coincide karte hain aur all-downstairs shorthand safe hai.

Why questions

Ek summed index bilkul do baar kyun appear karna chahiye, teen ya zyada baar nahi?

Summation factors ko pair karta hai,

\sum_{i} (this_{i}) (that_{i})

. Do copies ek unique pairing specify karte hain; teen copies factors group karne ka koi unique tarika nahi dete, isliye notation ambiguous hoga aur isiliye banned hai.

Free indices ki sankhya tumhe object ka order kyun batati hai (uski matrix rank nahi)?

Har free index independently

1 \dots n

par chalta hai, ek alag component generate karta hai: 0 → scalar, 1 → vector, 2 → matrix/order-2 Tensors. Yeh count tensor ka order hai (index slots ki sankhya), jo linear-algebra rank se unrelated hai (matrix ki independent rows ki sankhya) — same English word, alag concept.

Kronecker delta ko "substitution operator" kyun kaha jaata hai?

Kyunki

δ_{ij} a_{j} = a_{i}

: ek

δ

ko ek index ke against contract karna summed index delete karta hai aur doosre object ke index ko delta ke surviving wale se rename karta hai — yeh ek label ko doosre se swap karta hai.

Levi-Civita symbol naturally cross products aur Determinants encode kyun karta hai, jabki

δ

dot products encode karta hai?

ε

mein ek sign hoti hai jo swaps ke neeche flip karti hai (orientation), jo exactly wahi antisymmetry hai jo cross product ya determinant ko chahiye.

δ

symmetric hai aur bas indices match karta hai, jo symmetric Dot product ko chahiye.

ε_{ij k}

ke koi bhi do indices swap karne se uski sign kyun flip hoti hai, aur Vector calculus identities mein yeh useful kyun hai?

Koi do entries swap karna ek even permutation ko odd mein turn karta hai (aur vice versa),

+ 1 \leftrightarrow - 1

flip karta hai. Yeh antisymmetry tumhe indices ko cyclically reorder karne deti hai (

ε_{ij k} = ε_{k ij}

) taaki epsilon–delta identity apply karne se pehle shared index line up ho sake.

Ek single epsilon–delta identity

ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}

itne saare messy proofs ko kyun collapse kar sakti hai?

Yeh do orientation objects (mushkil, sign-laden) ke product ko plain index-matching

δ

's (aasaan substitutions) mein convert karta hai. Lagbhag har double-cross-product identity is ek line tak reduce hoti hai — jaise BAC–CAB rule.

Edge cases

1D mein,

δ

aur

ε

ka kya hota hai?

n = 1

ke saath har index sirf

1

ho sakta hai, isliye

δ_{ii} = δ_{11} = 1

aur koi bhi

ε

with

\geq 2

slots all zeros hai (ise non-zero hone ke liye distinct values chahiye). Ek 1D "vector" ka ek single component hota hai, isliye convention ordinary multiplication tak degenerate ho jaata hai.

2D mein, kya

ε_{ij k}

meaningful hai?

Nahi — 3-index

ε_{ij k}

intrinsically 3-dimensional hai. 2D mein tum two-index

ε_{ij}

use karte ho (with

ε_{12} = + 1, ε_{21} = - 1

). Symbol ke index count ko dimension

n

se match karo.

Convention

δ_{ij} a_{i} b_{j}

ke liye kya deta hai?

Delta

i = j

force karta hai, ise

a_{i} b_{i} = a \cdot b

tak collapse karta hai. Isliye contraction ke andar

δ_{ij}

do vectors ko unke dot product mein glue kar deta hai.

Kya

A_{ii}

ek scalar hai ya matrix?

Ek scalar — repeated

i

summed hai, deta hai

A_{11} + A_{22} + \dots + A_{nn}

A

का trace. Zero free indices bachte hain.

Agar ek term mein koi repeated index hi nahi hai, toh kya convention "kuch nahi kar raha"?

Sahi — bina repeated index ke koi implied sum nahi hai. Har index free hai, isliye term ek genuine multi-component object hai (jaise outer product

a_{i} b_{j}

), component-by-component evaluate hota hai.

i = j

hone par

ε_{ij k}

kya hai?

Zero. Koi bhi repeated index

ε = 0

bana deta hai, kyunki ek permutation koi value repeat nahi kar sakta. Yahi ek fact hai jo

a \times a = 0

force karta hai aur antisymmetry encode karta hai.

Kya ek valid equation ke ek side par free index ho sakta hai lekin doosri par nahi, jaise

c = a_{i} b_{i}

Haan yahan, kyunki dono sides ke zero free indices hain (

i

right par summed hai,

c

ek scalar hai). Rule sirf yeh hai ki free indices match karne chahiye — aur match karne ke liye koi nahi hain.

Kya

a_{i} b_{i}

versus

b_{i} a_{i}

mein factors ka order matter karta hai?

Nahi — har term numbers ka ordinary product hai, isliye

a_{i} b_{i} = b_{i} a_{i}

; sum commutative hai. (Order does matter karta hai

A_{ij} B_{j k}

ke liye, kyunki wahan shared-index positions define karte hain kaun row hai aur kaun column.)

Flat Euclidean space mein, kya ek index raise ya lower karne se koi numbers change hote hain?

Nahi —

g_{ij} = δ_{ij}

ke saath,

a^{i} = a_{i}

component-for-component, isliye upper/lower placement cosmetic hai. Yeh sirf tab matter karna shuru karta hai jab metric

δ_{ij}

se alag ho (curved space ya non-trivial signature).

Recall Ek-line self-test

Agar tum turant bol sako (a) ek term mein kitne free indices hain, (b) kaun sa index summed hai, aur (c) koi index illegally teen+ baar repeat to nahi ho raha — tum poora convention internalize kar chuke ho. "Twice = sum, Once = run, Thrice = crime." ::: Is page par har trap ka teen-word summary.

Connections

Parent: Einstein summation convention.
Traps liye gaye hain Dot product, Cross product, Matrix multiplication, Determinants, Tensors, aur Vector calculus identities se — helpers Kronecker delta aur Levi-Civita symbol ke saath.