Deviation probabilities for extremal eigenvalues of large Chiral non-Hermitian random matrices

Proof of Lemma 4.1.

The expression (4.1) reminds us to work on the integral ∫ c n , v ⁢ a n , v v + ∞ e - v ⁢ τ j ⁢ ( t ) ⁢ 𝑑 t and the factor. Utilize the Stirling formula

for z large enough to write

for j large enough, and this asymptotic still holds for j bounded since both Γ ⁢ ( j ) and ( j - 1 2 ) ⁢ log ⁡ j inherit the boundedness of j. When c n , v ⁢ a n , v v > x j , it follows from Lemma 2.5 that

By definition,

When v ≫ c n , v ⁢ a n , v or v = O ⁢ ( c n , v ⁢ a n , v ) , we see the unboundedness of

which is due to the fact c n , v ⁢ a n , v v ≥ 2 ⁢ n ⁢ a n , v → + ∞ . By conditions, we have

This implies that

For the case v ≪ a n , v ⁢ c n , v , we see v ≪ n and τ j ′ ⁢ ( c n , v ⁢ a n , v v ) = O ⁢ ( 1 ) . Therefore

and furthermore,

Also, when j satisfying c n , v ⁢ a n , v v > x j , Lemma 2.5 entails that

for any M > c n , v ⁢ a n , v v . Choose M = c n , v ⁢ ( 1 + a n , v ) v . Similarly, we see

For the term v ⁢ τ j ⁢ ( M ) - v ⁢ τ j ⁢ ( c n , v ⁢ a n , v v ) , since τ j ′ is increasing on [ x j , ∞ ) , we know

The analysis above shows that

which tends to the positive infinity since both v ⁢ τ j ′ ⁢ ( c n , v ⁢ a n , v v ) and c n , v v tend to the positive infinity. Hence, we claim that

Therefore,

(A.2)

for n large enough and for any j with c n , v ⁢ a n , v v > x j . Putting the expression

into (A.2), we have that

Now we suppose that c n , v ⁢ a n , v v ≤ x j . Lemma 2.5 implies the two inequalities

for any M > x j . Choose M = c n , v ⁢ ( 1 + a n , v ) v , which verifies the condition M > x j , since c n , v ⁢ a n , v v > x j . A similar argument leads again that log ⁡ ( v ⁢ τ j ′ ⁢ ( M ) ) = O ~ ⁢ ( log ⁡ n ) and

Therefore, we have that

Combining this asymptotic with the expression (A.1), we get that

for j such that c n , v ⁢ a n , v v ≤ x j . The second item of Lemma 2.5 says

which implies

Thus, by definition

Putting this back into (A.3) and combining like terms, we get

Proof of Lemma 4.2.

The argument will be similar to that of Lemma 4.1. For the second item, we only need to prove that

Lemma 2.5 guarantees that

and

once c n , v ⁢ a n , v v ≤ x j . Hence, it remains to prove that

Indeed, using the precise form of τ j , we have that

The condition that c n , v ⁢ a n , v ≤ v ⁢ x j ≤ 2 ⁢ j ⁢ ( j + v ) implies the following inequality:

Since v 2 + c n , v 2 ⁢ a n , v 2 4 ≥ v , it follows that

as j → ∞ . Hence,

It remains to prove that log ⁡ ( - v ⁢ τ j ′ ⁢ ( c n , v ⁢ a n , v v ) ) = O ~ ⁢ ( log ⁡ n ) . Recall

When c n , v ⁢ a n , v ≫ v , it follows from the fact c n , v ⁢ a n , v ≤ v ⁢ x j ≤ 2 ⁢ j ⁢ ( j + v ) that v ≪ j and j ≥ n ⁢ a n , v + o ⁢ ( n ⁢ a n , v ) . Thus

On the one hand, j ≤ n says

On the other hand, it is easy to see that

Consequently, we have the desired expression

The same argument works when v = O ⁢ ( c n , v ⁢ a n , v ) . Now we consider the last case that v ≫ c n , v ⁢ a n , v , which indicates the following lower bound of j:

Under this assumption, it holds that

as n → ∞ . The last limit is due to the fact that

Similarly, using the inequality n ⁢ a n , v ≤ j ⁢ v c n , v ⁢ a n , v ≤ n 2 ⁢ n ⁢ a n , v , we claim that

Now, we prove the first item, which means working on j with c n , v ⁢ a n , v v > x j . As for the second item of Lemma 3.1, it is enough to prove that

Indeed, Lemma 2.4 entails again that

for any 0 < M < x j . It is clear that

Choose M = j ⁢ ( j + 4 ⁢ v ) 2 ⁢ v . The remainder of the proof will be focused on

By definition, we have that

Hence,

Meanwhile, it follows from the definition of τ j and the fact x j > j ⁢ ( j + 4 ⁢ v ) 2 ⁢ v that

which tends to - ∞ as n → ∞ . The proof is then accomplished. ∎